spring 2015. horizontal alignment geometric elements of horizontal curves superelevation design...

HORIZONTAL ALIGNMENT

Spring 2015

Horizontal Alignment

Geometric Elements of Horizontal Curves

Superelevation Design

Transition or Spiral Curves

Sight Distance

PC PT

Simple Curve

Circular Curve

Tangent

Point of Curvature Point of Tangency

SC

ST

Curve with Spiral Transition

Circular Curve

Tangent

Tangent to Spiral

Spiral to Tangent

Spiral

TS

Spiral to Curve

CS

Curve to Spiral

Design Elements of Horizontal Curves

Deflection Angle

Deflection Angle

( )2IT RTan=

Also known as Δ


5729.58D

R=

Larger D = smaller Radius

100I

LD

=


E=External Distance

M=Length of Middle Ordinate


LC=Length of Long Cord

Basic Formulas

20.01

1 0.01 15

e f V

ef R

+=

-

Where,

e = superelevation

f = side friction factor

V = vehicle speed (mph)

R = radius of curve (ft)

2

0.0115

Vf e

R= -

Basic Formula that governs vehicle operation on a curve:

Basic Formulas

Where,

e = superelevation

f = side friction factor

V = vehicle speed (mph)

R = radius of curve (ft)

( )

2

minmax max15 0.01

VR

e f=

+

Minimum radius:

Minimum Radius with Limiting Values of “e” and “f”

2

maxd

Ve f

gR

Superelevation DesignDesirable superelevation:

for R > Rmin

Where,

V= design speed in ft/s or m/s

g = gravity (9.81 m/s2 or 32.2 ft/s2)

R = radius in ft or m

Various methods are available for determining the desirable superelevation, but the equation above offers a simple way to do it. The other methods are presented in the next few overheads.

Method 1:◦ Superelevation and side friction are directly

proportional to the inverse of the radius (straight relationship between 1/R=0 and 1/R =1/Rmin)

Method 2:◦ Side friction is such that a vehicle traveling

at the design speed has all the acceleration sustained by side friction on curves up to those requiring fmax

◦ Superelevation is introduced only after the maximum side friction is used

Methods for Estimating Desirable Superelevation

Method 3:◦ Superelevation is such that a vehicle traveling at the

design speed has all the lateral acceleration sustained by superelevation on curves up to those required by emax

◦ No side friction is provided on flat curves◦ May result in negative side friction

Method 4:◦ Same approach as Method 3, but use average running

speed rather than design speed◦ Uses speeds lower than design speed◦ Eliminate problems with negative side friction

Method 5:◦ Superelevation and side friction are in a curvilinear

relationship with the inverse of the radius of the curve, with values between those of methods 1 and 3

◦ Represents a practical distribution for superelevation over the range of curvature

◦ This is the method used for computing values shown in Exhibits 3-25 to 3-29

e = 0 ema

x

Reciprocal of Radius

Sid

e F

rictio

n F

acto

r Five Methods

fmax

M2 M1

M3

M5

M41/R

f

Important considerations:◦ Governed by four factors:

Climate conditions Terrain (flat, rolling, mountainous) Type of area (rural vs urban) Frequency of slow-moving vehicles

◦ Design should be consistent with driver expectancy

◦ Max 8% for snow/ice conditions◦ Max 12% low volume roads◦ Recurrent congestion: suggest lower than 6%

Design of Horizontal Alignment

Method 1Centerline

Method 2Inside Edge

Method 3Outside Edge

Method 4Straight Cross Slope

In overall sense, the method of rotation about the centerline (Method 1) is usually the most adaptable

Method 2 is usually used when drainage is a critical component in the design

In the end, an infinite number of profile arrangements are possible; they depend on drainage, aesthetic, topography among others

Which Method?

Median width

Pivot points

Example where pivot points are important

Bad design

Good design

15 ft to 60 ft

The superelevation transition consists of two components:◦ The superelevation runoff: length needed to

accomplish a change in outside-lane cross slope from zero (flat) to full superelevation

◦ The tangent runout: The length needed to accomplish a change in outside-lane cross slope rate to zero (flat)

Transition Design Control


Tangent Runout


Superelevation Runoff


http://techalive.mtu.edu/modules/module0003/Superelevation.htm

http://techalive.mtu.edu/modules/module0003/Superelevation.htm

Minimum Length ofSuperelevation Runoff


= relative gradient in previous overhead


Values for n1 and bw in equation

Minimum Length ofTangent Runout

See Exhibit 3-32 for values of Lt and Lr

Superelevation RunoffLocation: 1/3 on curve

Location: 2/3 on tangent

Superelevation Runoff

All motor vehicles follow a transition path as it enters or leaves a circular horizontal curve (adjust for increases in lateral acceleration)

Drivers can create their own path or highway engineers can use spiral transitional curves

The radius of a spiral varies from infinity at the tangent end to the radius of the circular curve at the end that adjoins the curve

Transition Curves -Spirals

Transition Curves -Spirals

Need to verify for maximum and minimum lengths

Transition Curves

Superelevation runoff should be accomplished on the

entire length of the spiral curve transition

Equation for tangent runout when Spirals are used:

The sight distance is measured from the centerline of the inside lane

Need to measure the middle-ordinate values (defined as M)

Values of M are given in Exhibit 3-53 Note: Now M is defined as HSO or Horizontal

sightline offset.

Sight distance on Horizontal Curve

Example ApplicationIncluded for your benefit

e = 0emax (for the design speed)


Sid

e F

rictio

n F

acto

rSelection of fdesign and edesign (Method 5)

fmax (for the design speed)

fdesign

1/R

f

e = 0emax


Sid

e F

rictio

n F

acto

rSelection of fdesign and

edesign

fmax

fdesign

Rf = V2/(gfmax)

Ro = V2/(gemax)

R0: f = 0, e = emax

Rmin = V2/[g(fmax + emax)]

1/R

f

e = 0

emax (for the design speed)


Sid

e F

rictio

n F

acto

rSelection of fdesign and

edesign

fmax (for the design speed)

fdesign

fdesign = α(1/R)+β(1/R)2

α = fmaxRmin[1-{Rmin/(R0-Rmin)}]

β = fmaxRmin3/(R0-Rmin)

1/R

f

Superelevation Design for High Speed Rural and Urban Highways

Example:

Design Speed: 100 km/h

fmax = 0.128

emax = 0.06

Question?

What should be the design friction factor and design superelevation for a curve with a radius of 600 m?

1. Compute Rf, R0, and Rmin:

Rf = V2/(gfmax) = 27.782 / (9.81 x 0.128) = 615 m

R0 = V2/(gemax) = 27.782 / (9.81 x 0.06) = 1311 m

Rmin = V2/[g(fmax + emax)] = 27.782 / [9.81(0.128+0.06)]

Rmin = 418 m

e = 0

emax = 0.06 Sid

e F

rictio

n F

acto

rSelection of fdesign and edesign

(example)

fmax = 0.128

fdesign

1 / 1311 1 / 615 1 / 418 1/R

f

2. Compute α and β:

α = 0.128 x 418 x [1 – 418 / (1311 – 418) ] = 28.45 m

β = 0.128 x 4183 / (1311 – 418) = 10502 m2

3. Compute fdesign and edesign :

First, estimate the right-hand side of equation for designing superelevation

e + f = V2/(gR) = 27.782 / (9.81 x 600) = 0.131

Then,

fdesign = 28.45 / 600 + 10502 / 6002 = 0.076

edesign = 0.131 – 0.076 = 0.055 (< emax = 0.06)

e = 0

emax = 0.06 Sid

e F

rictio

n F

acto

r

fmax = 0.128

fdesign

1 / 1311 1 / 615 1 / 418

1 / 600

0.076

Selection of fdesign and edesign

(example)

1/R

f

Selection of fdesign and edesign

(example)

R=600 ft

spring 2015. horizontal alignment geometric elements of horizontal curves superelevation design...

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