TTE3810 Geometric Design Part One_Fall 2013.pdf

8/16/2019 TTE3810 Geometric Design Part One_Fall 2013.pdf

http://slidepdf.com/reader/full/tte3810-geometric-design-part-onefall-2013pdf 1/60

TTE3810 CLASS

HIGHWAY GEOMETRIC DESIGN (PART-I-)

Prepared By

Dr. Haitham Al-Deek, P.E.

Professor of EngineeringDepartment of Civil and Environmental Engineering

University of Central Florida

Orlando, FL 32816-2450Phone: (407) 823-2988Fax: (407) 823-3315

E-Mail: [email protected]

mailto:[email protected]






Professor H. Al-Deek, Ph.D., P.E. Class TTE3810 Highway Geometric Design, Part-I

Phone 407-823-2988 [email protected]

Table of Contents

Page Number

I- Design Features 1

II- Elements of Highway Geometric Design 1

1. Design Speed 4

2. Horizontal Alignment 6

2.1 Simple Curve 6

2.2 Review of Curvilinear Motion and Derivation of 13 Superelevation Equation

2.3 Superelevation 16

2.4 Transition Curves (Spirals) 27

3. Vertical Alignment 32

4. Sight Distance 44

4.1 Types of Sight Distance 44

4.2 Sight Distance on Horizontal Curves 46

4.3 Sight Distance on Vertical Curves 51

Appendix A: Metric Tables for Sight Distance on Vertical Curves 55








- 1 -

I.

DESIGN FEATURES

Geometric Design

Highway Alignment X-Section Design

x

z

y

x - Latitude y - Departure (Positive y ⇒ N)

z - Elevation above mean sea level

pts. along route are called stations( the distance in feet from somereference point)

At right angle to direction of alignment

Describes widths, clearances and slopes...

The essential design features of a roadway are its location and its cross section. In the horizontal

plane, the locations of points are referenced to a coordinate system in which the positive y-axis is

north and the positive x-axis is east.

Points along the route are identified by stations. A station is defined as the distance in feet or

meter from some reference point, commonly the beginning point of the roadway. The location

of points in the vertical plane (or along z-axis) is given as the elevation above the mean sea level.

The cross section of a roadway is described by its dimensions at the right angle to the direction

of the alignment, including widths, clearances, slopes, and so on.

II. ELEMENTS OF HIGHWAY GEOMETRIC DESIGN

1. Design Speed

2. Horizontal Alignment

2.1. Simple Curve

2.2. Superelevation (e)








- 2 -

2.3. Transition Curves (Spirals)

3. Vertical Alignment

4. Sight Distance

4.1. Types of Sight Distance4.2. Sight Distance on Horizontal Curves

4.3. Sight Distance on Vertical Curves

5. Earthwork (covered in the “Surveying” course).

6. Cross Section Elements

7. Intersections and Interchanges

7.1. Intersection at Grade.

7.2. Grade Separations and Interchanges

The relationship of traffic to highway design

Traffic elements that influence highway design include the following:

• Average Daily Traffic (ADT)

• Design Hourly Volume (DHV)

• Directional Distribution (D)

• Percentage of Trucks (T)

• Design Speed

Selection of the Design Hourly Volume (DHV)

Average Annual Daily Traffic "AADT" (vehicles per day) is not appropriate to use in the

geometric design of highways. Traffic engineers use hourly traffic volumes for design. Which

hourly traffic volume should be used?

The maximum peak hourly traffic volume during the year? Or

The average hourly traffic volume during the year ?

Answer: None!!








- 3 -

5

10

15

20

30

25

H o u r l y V o l u m e a s p e r c e n t o f A A D T

Source: Minnesota Department of Transportation

100 300200 600500400 1000900800700 1100

EXHIBIT 8-8. RANKED HOURLY VOLUMES

Recreational Access Route MN 169

Main Rural Route I-35

Urban Circumferential Freeway I-494

Urban Radial Freeway I-35E

Hour Rank

Source: HCM2000 and AASHTO 2001

24

14

10

30 100

H o u r l y V o l u m e

Number of hours in one year with hourly volume

greater than that shown

AADT

The 30th highest hourly volume is used as the design hourly volume. This means that the

designer is willing to tolerate subjecting the public (travelers) to congested conditions for 29

hours during the year (this is only 0.3% of all hours during the year).

This Volume is referred to as DHV.








- 4 -

DHV K ADT = ×

K does not vary much from year to year!

8 – 12% Urban highways

K= 10-15% Suburban highways

12-18% Rural highways

FDOT (1995) uses different values. See below:

Area Type K-FactorUrbanized 0.091

Urban 0.093Transitioning/Urban 0.093Rural Developed 0.095Rural Undeveloped 0.100

Note that DHV is the sum of hourly volume for both directions. To find the “Directional Design

Hourly Volume” or DDHV we need to multiply the DHV by a distribution factor called D to

split traffic volume between the two directions, or use ADT to calculate DDVH as follows:

DDHV K D ADT = × ×

Where,

D = Directional split factor, which splits traffic into the heavy direction. So, D might be as high

as 0.7 or 70% in the heavy direction in rural highways during design hour, and the off-peak

direction carries only 30% of the traffic. Normally, the peak/off-peak volume switches

directions between morning and evening peak hours. This means that one direction will be

heavy in the morning while the opposite direction will be heavy in the evening.

1. DESIGN SPEED

The elements of highway design are influenced by the design speed. The selection of the

appropriate design speed depends on:

Highway type (rural vs. urban, interstate, local)








- 5 -

Traffic characteristics (volume and composition).

Vehicle characteristics (trucks, passenger cars).

Cost of land.

Topography (level, rolling, mountainous).








- 6 -

2.

HORIZONTAL ALIGNMENT

Horizontal curves are curves in plan used to provide change in direction of the center line of the

road.

2.1 Simple Curve

A simple curve is a circular joining two tangents.

x

y

φ

Pl = arc

lengthPC to P

PC

R θ

θ

PC PT

PI

P

R R

R

100'

D

D = central angle

for 100' arc

∆

2∆

C

M

4∆

Back

Tangent Forward

Tangent

L =arc

length

PC to PT

∆

αT E

( ,R)

VARIABLES

PC = Point of curvature (Beginning of curve) PI = Point of intersection

PT = Point of tangency (End of curve) D = Degree of curvature (1)

PI = Point of intersection R = Radius of curve

∆ = Central angle (Intersection angle) E = External distance

L = Length of curve (PC to PT) M = Middle ordinate

l = Length of arc (PC to P) C = Chord Length

θ = Central angle for arc length l

T = Tangent length (PC to PI & PT to PI)

φ = Deflection angle at PC between tangent and chord for P

α = Deflection angle at PI between tangent and line from PI to P

x = Tangent distance from PC to P

y = Tangent offset P

Note: (1) This variable is used only for curve definition in traditional US units

Properties of a simple circular curve








- 8 -

Example 1:

A circular horizontal curve has an intersection angle ∆ = 23o 42' right and degree of curve

D = 1o 45′ (chord definition). Compute the tangent distance, curve length, and station of the

point of the tangency PT . The station of point of intersection PI = 27+85.50

Solution to Example 1:

Using the degree of curve (chord definition)

50 50R 3274.1 ft

1 45sin sin

2 2

D= = =

′°

23 42T = R * tan 3274.17* tan 686.99 ft

2 2

′∆ ° = =

2 R 2 *3274.17*23.7L 1354.34 ft

360 360

Π ∆ Π °= = =

PI station = 27+85.50 = 2785.50 ft

PC station = PI station – back tangent distance = 2785.50 - 686.99 = 2098.51 ft = 20+98.51

PT station = PC station + curve length = 2098.51 + 1354.34 = 3452.85 ft = 34+52.85

Example 2:

Two tangents are to be connected by a circular curve with D = 4°. The tangents intersect at

station (28+42.83) with ∆ = 52°30′. Find R, L, T, E, stations of PC and PT.


5729.6R 1432.4 ft

4= =

PCPT

PI ∆








- 9 -

( ) ( )2*3.1415

L 1432.4 52.5 1312.5 ft360

= =

52.5T 1432.4* tan 706.4 ft

2

= =

52.5E 706.4* tan 164.7 ft

4

= =

PC station = (28+42.83) – (7+06.40) = (21+36.43)

PT station = (21+36.43)+(13+12.5) = (34+48.93)

Example 3:

The two tangents described below intersect 2000 ft beyond station 10+00, back tangent

N45°00′00′′ W, forward tangent N 15°00′00′′ E. It is desired to design a 3000 ft radius

horizontal curve between the two tangents. Determine degree of curve (arc. def.), the tangent

distance, length of curve, station of PC, and station of PT.

PC

PT

PI

∆

10+00

2000 ft

45°15°

N

Not to scale

R=3000 ft

N 15°00'00"E

N 45°00'00"W


∆ = 15o 00′00′′ + 45o

00′00′′ = 60o 00′00′′

R = 3000 ft








- 10 -

100*180D = 1.91

R = °

π------- (arc. def) or

5729.6D =

R

60°T = R*tan = 3000*tan = 1732.05 ft

2 2

∆

2 R 2 *3000*60°L = = = 3141.59 ft

360 360

Π ∆ Π

Station of PI = 1000 + 2000 = 3000 ft = 30+00

Station of PC = Station of PI - Tangent length

= 3000 - 1732.05 = 1267.95 ft = 12+67.95

Station of PT = Station of PC + Length of curve

= 1267.95 + 3141.59 = 4409.54 ft = 44+09.54

Example 4:

Given highway curve of 10°

PC PT

PI

Station 50+00

Δ= 200 34'

D = 100

Required

a) Find T, station of PC and PT, length of curve, length of long chord, and middle ordinate M.

b) Calculate the chord length between stations 49+50 and 50+50.

c) With forward tangent moved out 2 ft and PC held, what is the length of the new curve? See

figure below.








- 12 -

Find R new ?

Toriginal = 103.95 ft

2X = = 5.69 ft

sin 20° 34′

Tnew = 103.95 + X = 103.95 + 5.69 = 109.64 ft

R original = 572.96’

new

109.64R = = 604.31 ft

20.5667tan

2

newL = 604.31* *20.5667 = 216.92 ft180

Π








- 13 -

2.2 Review of Curvilinear Motion and Derivation of Superelevation

Equation

Curvilinear Motion

an

at

V

R

C

Vehicle’s acceleration Horizontal (tangential) at

Normal an

2

, if = constant => 0t t

n

dva v a

dt

va

R

= =

=

Also, for a circular path R is fixed. Take a horizontal cross-section AA for a highway.

Direction of velocity is tangent to the

path.








- 14 -

R

C

A A

t

n

A AF

W

R

Lateral effect

Side friction resists the tendency of vehicle to slide outward in the circular path. To minimize

this tendency, highway design provides for superelevation of the section of the roadway => the

cross-section is tilted by an angle θ.

F

W

R

θ

Superelevation is used to counteract the effect of the outward acting centrifugal force. The outer

edge of the road should be raised with respect to the inner edge producing a traverse slope

throughout the whole length of the horizontal curve.








- 15 -

2

cosW vg R

θ

2W vg R

θ

θF

Fx

y

2

sinW vg R θ

W cos θ

W

W sin θ

The components of the vehicle’s weight along the tilted pavement surface (X-axis) resist the

sliding tendency of the vehicle. Friction + Weight component resist this tendency of sliding

outward.

f = friction factor F= f. N

( )

( )

2

2

cos sin 1

from X-axis => sin cos 2

w vF f w

g R

w vF w

g R

θ θ

θ θ

= +

+ =

From (1) and (2)

( )

( )

( )

2 2

2 2

222 2 2

2

cos sin sin cos 3

divide by cos

tan tan

Let tan be = e, f.e 01.47

32.16 14.88 15

15

Note: V is in mph, R i

wf v w vwf w

g R g R

w

v v f f

gR gR

V v V V e f

gR R R R

V e f

R

θ θ θ θ

θ

θ θ

θ

+ + =

+ + =

≈

+ = = = ≈

+ =

s in ft.








- 16 -

2.3 Superelevation (e)

Superelevation (e) is defined as the ratio of the height of the outer edge to the horizontal width of

the road.

θ h

P.C.

P.T.

outer "0"

inner

"i"

h = e

L

L

2 cosW vg R θ

2W vg R

θ

θF

Fx

y

2

sinW vg R

θ

W cos θ

W

W sin θ

Analysis of the forces acting on the vehicle as it moves on a circular curve results the following

formula for calculating the minimum design radius :

R 15

Vf e

2

=+ (traditional U.S. units)

Where

e = superelevation rate

f = friction factor

V = design speed in mph

R = minimum radius, ft

R 127

Vf e

2

=+ (metric system)

Where

e = superelevation rate

f = friction factor

V = design speed in kph

R = minimum radius, m

This equation shows that the effect of centrifugal force is resisted partly by superelevation(e) and

partly by friction between tires and road surface(f). From this equation, it can be seen that as the

summation of (e+f) increases, the required radius R decreases. So, there is a minimum allowable

R and a maximum allowable (e and f).








- 17 -

emax : According to AASHTO "Green Book", the maximum allowable design value for "e" is

0.12 ft/ft. Present practice suggests emax = 0.10 ft/ft. In locations where snow and ice

conditions occur, emax = 0.08.

f max: The maximum design values for "f" depend on design speed. Value of "f" ranges

between 0.17 and 0.10 for design speed range of 20 to 70 mph.

Rmin: For a given design speed, R min and Dmax (max. safe degree of curvature) can be

calculated from the maximum rate of superelevation "e" and the side friction factor (f)

as follows:

)f e(15

VR

.max

2

.min +=

which is equivalent to:

2

max

.maxV

)f e(900,85D

+=

where,

R min.= minimum safe radius "R" of the horizontal curve.

Dmax= maximum safe degree of the horizontal curve.

Question: Why do we set maximum limits on superelevation e?

Answer:

There are three reasons:

1) To prevent slow moving vehicles from sliding to the inside of the curve,

2) In urban areas, to keep parking lanes relatively level; and

3) To keep the difference in slope between the roadway and any streets or driveways that

intersect it within reasonable bounds.








- 18 -

Example 5:

• Calculate the maximum degree of curve and R min of a simple circular curve with an externalangle of 1000. The design speed is 50 mph, the corresponding value of f max is 0.14 and emax =0.10 ?

• Calculate edesign for a curve that has a radius of 800’ ? (given above f max)


( ) ( )

( )

( )

2 2

min

min

0

2

0

2 2

2

50694.4'

15 15 0.1 0.14

695'

85,900 0.1 0.148.2464

50,

5729.588.24

695

Note: External angle does not enter these calculations.

800'

15 15

50 0.14 0.0715 800

design

V R

e f

R

D

or

D

R

V V e f e f

R R

e

= = =+ +

≅

+= =

= =

= ⇒

+ = ⇒ = −

= − =








- 19 -

Example 6

Example of Minimum Raduis Data for Limiting Values of e and f :

Rural Highways and High-Speed Urban Streets

DesignSpeed

(km/hr)

Maximume

Maximumf

Total(e + f)

CalculatedRadius

(m)

RoundedRadius

(m)120 0.06 0.09 0.15 755.9 755

30 0.08 0.17 0.25 28.3 30

40 0.08 0.17 0.25 50.4 50

50 0.08 0.16 0.24 82.0 80

60 0.08 0.15 0.23 123.2 125

70 0.08 0.14 0.22 175.4 175

80 0.08 0.14 0.22 229.1 230

90 0.08 0.13 0.21 303.7 305

100 0.08 0.12 0.20 393.7 395

110 0.08 0.11 0.19 501.5 500

120 0.08 0.09 0.17 667.0 665

30 0.10 0.17 0.27 26.2 2540 0.10 0.17 0.27 46.7 45

50 0.10 0.16 0.26 75.7 75

60 0.10 0.15 0.25 113.4 115

70 0.10 0.14 0.24 160.8 160

80 0.10 0.14 0.24 210.0 210

90 0.10 0.13 0.23 277.3 275

100 0.10 0.12 0.22 357.9 360

110 0.10 0.11 0.21 453.7 455

120 0.10 0.09 0.19 596.8 595

30 0.12 0.17 0.29 24.4 25

40 0.12 0.17 0.29 43.4 45

50 0.12 0.16 0.28 70.3 7060 0.12 0.15 0.27 105.0 105

70 0.12 0.14 0.26 148.4 150

80 0.12 0.14 0.26 193.8 195

90 0.12 0.13 0.25 255.1 255

100 0.12 0.12 0.24 328.1 330

110 0.12 0.11 0.23 414.2 415

120 0.12 0.09 0.21 539.9 540Source: A Policy on Geometric Design of Highways and Streets , American Association of State Highway andTransportation Officials, Washington, DC, 1994.

Given the table above, calculate the superelevation rates for a roadway with a design speed of 90

km/hr that has a wide range of curve radii, that is, R has values of 585, 440, 350, 295, and 250 m.

(These values correspond to degrees of curve D of 3, 4, 5, 6, and 7.) Use emax = 0.10. Compare

the results with those obtained in the following figure.








- 20 -


By the above table, use a maximum side friction factor of 0.13.

By the equation2

127

V e f

R+ = , the following values of e are calculated.

R (m) Computed value of e Recommended design e, fromfigure below

585 -0.021 0.070

440 +0.015 0.083

350 +0.052 0.094

295 +0.086 0.099

250 +0.125 Exceeds emax

Superelevation of Railway and Transit Guideway Curves

7000

5000

3000

20001500

1000

700

300

500

150

200

100

30

50

70

R a d i u s o f c u r v e ,

R ( m

)

e m a x

= 1 0 . 0

%

Superelevation rate, e (%)

0.0 10.09.08.07.06.05.04.03.02.01.0 11.0

30

40

50

100

80

70

60

90

110120

V, km/h

Design superelevation rates, emax

= 0.10. (Source: A Policy on Geometric Design of

Highways and Streets, copyright 1994, American Association of State Highway and Transportation

Officials, Washington, DC. Used by permission.)

Discussion:

For the sharpest curve (the 250-m curve), the combination of maximum superelevation rate and

the maximum side friction factor is insufficient to offset the centrifugal force. This curve is too

sharp for the given design speed and maximum superelevation rate and would be unsuitable forthe stated conditions.

At the other extreme, for the 585-m curve, a negative value of e was computed. Along this

curve, all of the centrifugal force could be offset without exceeding the recommended f max value

of 0.13, even with zero superelevation. The AASHTO favors a distribution of superelevation








- 21 -

that provides a logical relation between the side friction factor and applied superelevation rate

and recommends a positive amount of superelevation for the flattest curve.

The minimum radius of curve for a given design speed can be determined from the maximum

rate of superelevation and the side friction factor. The minimum comfortable radius R can becalculated from the following:

( )

2

min

max

VR

127 e f =

+ (metric system)

( )

2

min

max

VR

15 e f =

+ (traditional U.S. units)

The relationship between superelevation and minimum radius of a curve is shown in the previous

table for selected metric design speeds.








- 22 -

Superelevation Design:

Types of X-Sections

P.T. P.C.

P

Normal Crown (nc)

P

Adverse Crown

Removed Cross Fall

efull

efull (design)

Crown runoff

Direction

of tilting

P

w

Superelevation runoff: Transition of the x-section from adverse crown removed on the tangent

to a fully superelevated pavement (edesign) on the curve.

Current Practice:

P

efull

Y

Y

P.T.

Crown

runoff -P

ww

Lc

o

L3

2L3

P

P

P'








- 23 -

w: width of section being rotated (ft), Lc to edge

L: Length of superelevation runoff (ft)

e: Superelevation rate (ft/ft)

r: slope ratio of outer edge of pavement in relation to Lc

Superelevation profile

Transition of the x-section from the normal crown on the tangent to fully superelevated

pavements (e design) on the curve.

crown

runoff

Y Y

23 L 1

3 L

Outer

edge

CLLevel

Inner

edge

on tangent on curve

PP

P'e

e fullCross

Fall

Normal

Crown

P . C . o r P . T .

E l e v a t i o n ( f t )

Adverse Crown

Removed

The Superelevation transition (runoff) for simple circular curves is located partly on the

tangent and partly on the curve. Standard practice assigns 2/3 of the length on tangent. This

rule will only apply to simple curves. A crown runoff is also provided before the

superelevation runoff.

r is defined as the relative change of slope between the edges of the section being rotated and

the centerline.








- 24 -

r

ewL

⋅=

where :

w: width of section being rotated (ft)

L: length of superelevation runoff (ft)

e: superelevation rate (ft/ft)

r: slope ratio of outer edge of pavement in relation to centerline. Typical values of "r" are

1:200 (or 0.5%).

Example 7:

Draw a superelevation diagram (rotation around centerline) for a right-turn horizontal curve of a

2 lane 2 way rural highway. Given :

Lane width = 12.5 ft Normal crown slope (p) = 2% PC station = 101+00

Superelevation (e) = 6 % , "r" = 0.5%

The length of the horizontal curve = 200 ft


(see following two figures for superelevation profiles)








Example 7 Superelevation Profile








- 28 -

2.4 Transition Curves (Spirals)

Transition curves serve the purpose of providing a gradual change from the tangent section to the

circular curve and vice versa. A vehicle that enters a circular curve with transition travels

smoothly and naturally along a curve that gradually changes from straight line (infinite radius) to

a radius of some finite value, R, which is maintained throughout the length of the circular curve.

The most commonly used transition curve is the spiral. Spiral transitions normally are used only

on high-volume highways where the degree of curvature exceeds about 3°.

∆

∆c θ s

θ s

R c

S T

CSSC

L C

T S

K

L. T.

T S

PI

S. T.

θ s

E s

∆

PC

P φ c

PI - PT OF INTERSECTIONTS - TANGENT SPIRAL

SC - SPIRAL CURVE

CS - CURVE SPIRALST - SPIRAL TANGENT

RC - RADIUS OF CIRCULAR CURVE

L.T.- LONG TANGENT

S.T. SHORT TANGENT

LS - LENGTH OF SPIRAL

L - LENGTH TO ANY PT ON SPIRAL FROM TS

LC - CHORD DISTANCE - TS TO SC

S - CENTRAL ANGLE OF SPIRAL, "SPIRAL ANGLE"

- CENTRAL ANGLE OF ARC L

- DEFL. ANGLE FOR ANY PT. ON SPIRAL

FORMULAS3V

RCL = 1.6 , WHERE V = DESIGN SPEED

s

C - DEFL ANGLE FOR SC - INTERSECTION ANGLE,

TOTAL CENTRAL ANGLEC - CEN ANGLE OF CIRCULAR ARC

LC - LENGTH OF CIRCULAR ARCDC - DEGREE OF CURVE OF

SHIFTED CIRCLE

ES - EXTERNAL DIST FROM PI

TO CIRCULAR CURVE

P - OFFSET DIST OF CIRCULAR

CURVE PRODUCED

K - DIST FROM TS TO PC OF

CIRC CURVE PRODUCED

o

S CL D

S 200

C S

θ =

Δ =Δ 2θ−

( )

( )

( )( )

s

S

S

ΔS C 2

ΔS C 2

2O L3 L

100 Δ-2θ

C D

E = R +P exsec + P

T = R +P tan + k

=

L =

φ

Equations of Spirals

ss

c

Lθ = (radians)

2R or s c

s

L Dθ = (degrees)

200

C S2∆ = ∆ − θ

( )S

C

c

100 Δ 2θL =

D

−








- 29 -

Where :

LS = length of spiral, ft θS = spiral angle

R c = Radius of circular curve Dc = degree of circular curve

Lc = length of circular curve, ft ∆C = circular angleθ = central angle of arc L Φ = deflection angle for any point on the spiral

ΦC = deflection angle for SC ∆ = intersection angle, total central angle

TS = tangent distance.

Length of transition curve

When spiral transitions are used, superelevation should be attained within the limits of the

transition, (i.e., T.S-S.C , C.S.- S.T.) segments. Note that the two third L rule does not apply

here as in the simple curve. The minimum length of the transition curve is given as

C

3

sR

V6.1L = (traditional U.S. units)

Where

V : design speed in mph

R c : radius in ft.

Ls : transition length in ft.

Example 8:

For a highway of design speed 44 mph, two tangents intersect at station (127+23.68) with

∆ = 54°. These tangents are to be connected by spiral curves in conjunction with a simple

circular curve of D = Dc = 4°. (P = 0.29, K = 50.00)

Find spiral length, spiral central angle, circular curve central angle, circular curve length, tangent

distance, and stations for the entire curve.








- 30 -

T S

P

SC CS

ES

∆C

θs

θs

S T

T S

K

PI (127+23.68)

∆ = 54°

L C

∆

R C

LS L

S

LC


( )33

s

1.58 441.58VL = 94 ft

5729.6R

4

= =

Round up to the nearest 50 feet Choose Ls = 100 ft. Question: Why do we round up?

s cs

L D 100*4 = = = 2

200 200θ

∆c + 2θs = ∆ = 54°, ∆c = 50°

cc

c

100 100*50L = = = 1250 ft

D 4

∆

sT = (R+P) tan + K = 780 ft2

∆

Station of TS = (127+23.68) – (7+80) = (119+43.68)








- 31 -

Station of SC = (119+43.68) + (1+00) = (120+43.68)

Station of CS = (120+43.68) + (12+50) = (132+93.68)

Station of ST = (132+93.68) + (1+00) = (133+93.68)

In the following page we show a superelevation highway figure that includes both simple

circular curve and transition curve.








- 32 -








- 35 -

Vertical Curve Derivations

( )

2

2

2

2

1 2

0

1

2 11 2

To make life easier use constant = r

. . constant

To have a smooth ride and

1

What is r ?

x x L

x L

d y

dx

dy d y

dx r dx rxdx dx

dy dyg g

dx dx

dyg rx

dx

dy g gg rL g r dx L

→ →

=

⊕ → =

= = = +

⊕ → → →

⇒ = +

−= + = ⇒ =

∫ ∫

( )

( )

2 11

2

1 1

2

Combine (1) and (2)

grade at any point x

Elevation constant2

dy g gg x

dx L

dy

dx

dy rx y dx g rx dx g x

dx

−= +

=

= = + = + +

↑

∫ ∫

( )

( ) ( )

0

2

0 1

1 2

= = Elevation at BVC

32

Units: Any consistent set of units ft, meters...etc

g ,g are dimens

y

rx y x y g x= + +

• ionless

r ? must have compatable units as x and y.

1r units are reciprocal feet i.e. if x and y are in ft.

ft Practicing engineers use % for grades

e.g., g=0.03 3% rise of 3 ft in one stat

• →

•

→ → ( )ion or 100'

r=0.6% per station instead of 0.00006/ft, therefore Eq. (3) :becomes








- 36 -

( )

( )

0

0

1 2 1 2

1 2 1 2

G (G G )

100 200L

G (G G )Elev. of Point P on the Curve Elev. of BVC 4

100 200L

, :Where

y x y x x

x x

y is the BVC and this equation can be stated as

−= + +

−= + +

Equations of vertical curves

VPC or BVC VPT or EVC

Sta. of BVC = Sta. PVI -2

L

Elev. of BVC = Elev. of PVI -1G L

100 2

Sta. of EVC = Sta. PVI +2

L = Sta. VPC + L

Elev. of EVC = Elev. of PVI2G L

100 2

For any point "p" at a distance x from "BVC":

Sta. of P = Sta. of BVC + x

1 2 1 2G (G G )Elev. of P Elev. of BVC

100 200L x x

−= + +

Grade at P = G1 + AL

x

For any two points:

2

2 2 21

1 2 E

L

2

y y

x x

′ ′= =

High or Low Point on the Curve

The high (or low) point on the vertical curve is located at a distance XHP in case of crest curve

and XLP in case of sag curve. For simplicity we refer to this as X. The high (or low) point on the

vertical curve is located at a distance X.

1

1 2

LGX=

G -G

Where : L is the length of vertical curve in ft., X in ft.








- 37 -

BVC EVC

PI

L/2

x

High point

Crest

BVC EVC

PI

L/2

x

Low point

Sag

1

1 2

is in stations x is in stations isin ft x is in ft

LG x

G G

If L If L

=−

⇒⇒

Hints on Vertical Curves

( )2 1

The distance from the P.I to the middle of the parabolic curve:

800

where - , difference in grade in percent e.g., 0.03=3%

AL E

A G G

=

=

[ ]

Note that A is +ve if "sag" A is -ve if "crest"

+ve for "sag"E and L are in ft. and E is

-ve for "crest"

E is the distance that connects the midpoint of the curve above or below the P.I

to the P.I








- 38 -

E

E

Sag (E is below the

curve and above PI

Crest (E is above the curve

and below PI)

Offsets (Two Types)

2

1) Offset: is between curve and the gradient passing through the BVC. This is given by:

' 4

E, x, and L are all in ft and y' is in ft.

x y E

L

=

2

22 1

or, offset ' 4 800 200

x and L are in ft.

AL x G G

y x L L

−

= =

BVC

EVC

PI

Offset y'

( )

2

22 1

If x, L are in stations and E is in ft. Use same equation ' 4

If x, L are in stations and ' is calculated as:

' y' will be in stations not ft2

x y E

L

y

G G y x

L

⇒ =

−= ⇒

2) Offset: distance between curve and both gradients that pass through BVC and EVC.








- 39 -

BVC

EVC

PIy-

y'

L/2 L/2

( )

( )

2

2

4 ,2

'

4 - - ,2 2100

x L E x L

y x x A L L E x x L

≤

= ≥

Example 9:

A 600 ft. sag vertical curve has a -4% grade meeting a +4% grade. How long must a new vertical

curve be in order to retain the same grade rates and to raise the elevation at the center of the

curve by 2 ft.


( ) ( )( )(4-(-4)) 600 8 600AL (G2-G1)LE= = = = = 6

800 800 800 800′

new E = 6+2 = 8

( )8 L8

800=

Lnew = 800 ft








- 40 -

Example 10:

An ascending 5% grade and a descending 3% grade are joined by an 800 ft VC. Elevation at Sta

24+00 on the 5% grade is 1776.39 ft and elevation at Sta 36+00 on the 3% grade is 1780.39 ft.

Find the station of the BVC.


PVI

- 3%

EVC400'400'

d2

Sta 36+00

Elev 1780.39

Sta 24+00

Elev 1776.39

d1

5%BVC

d 1 + d 2 = 3600 - 2400

d 1 + d 2 = 1200 (1)

1776.39 + 0.05(d 1) - 0.03(d 2) = 1780.39

0.05(d 1) - 0.03(d 2) = 4 (2) substitute (1) into (2)

0.05(1200 – d 2) - 0.03(d 2) = 4

60 - 0.08 d 2 = 4 d 2 = 700 ft d 1 = 500 ftSta of P.I. = 2400 + d 1 = 2400 + 500 = 2900 ft = 29+00

Sta of BVC = 2900 - 400 = 2500 ft = 25+00








- 41 -

Example 11:

Two tangent highway alignments intersect in plan view as shown in the sketch below. Each

alignment consists of a vertical curve in the vicinity of the intersection point. The two vertical

curves data are given in the table below. Find the vertical clearance between the highwayalignments at point 'P'.

A l i g n m e n t # 1

A l i g

n m e n t #

2

Point 'P'

Sta 80+25 on Alignment #1

Sta 130+50 on Alignment #2

Data Alignment #1 Alignment #2

PVI Station 84+00 130+00

PVI Elevation (ft) 175.00 145.00

Length of Vertical Curve 1200 ft 800 ft

Tangent grades G1

G2

+2 %

-3 %

-1 %

+2 %








- 43 -

AL (G2-G1)L (2-(-1)(800) (3)(800)E = = = = = 3

800 800 800 800′

22

' 3

(13400 13050) 800

2

y=

−

y′ at Sta. 130+50 = 2.297′

Elev of 130+50 on tangent = 145 + (0.02) (13050-13000) = 146′

Elev of 130+50 on curve = 146 +2.297 = 148.297′

Clearance is the difference between Elevation as in Alignment#1 and elevation as in

Alignment#2 . Clearance = 166.445 - 148.297 = 18.148′

Example 12:

The grade line is being developed for a six-lane freeway with a design speed of 70 mph. A rising

2.0% grade meets a falling grade of 1.5% at station 100+00. The elevation of the P.I. of the two

grades is 98.00 ft.

a. Using vertical curve length of 1400 ft, compute the middle ordinate of the vertical curve.

b. For each station ,(i.e., 93+00, 94+00, etc…), calculate VC elevation.


PVI

Sta 100+00

Elev 98.00

EVC

Sta 107+00

-1.5%+2%BVC

Sta 93+00

S t a

9 4 + 0 0

S t a

9 5 + 0 0

S t a

9 6 + 0 0

S t a

9 7 + 0 0

S t a

9 8 + 0 0

S t a

9 9 + 0 0

S t a

1 0 0 + 0 C

S t a

1 0 1 + 0 C

S t a

1 0 2 + 0 C

S t a

1 0 3 + 0 C

S t a

1 0 4 + 0 C

S t a

1 0 5 + 0 C

S t a

1 0 6 + 0 C

E

L = 1400'

AL (G2-G1)L (-1.5-2)(1400) (-3.5)(1400)E = = = = = -6.125

800 800 800 800′








- 44 -

( )2 2

E offset=

( )L

2

x Offset =

( )

2 22

2 2

E 6.1250.0000125

(700)1400

2

x x x

−= = −

Station Tangent Elevation VC Offset VC ElevationBVC 93+00 84.00 0.000 84.0094+00 86.00 - 0.125 85.87595+00 88.00 - 0.500 87.50096+00 90.00 - 1.125 88.87597+00 92.00 - 2.000 90.00098+00 94.00 - 3.125 90.87599+00 96.00 - 4.500 91.500P.I. 100+00 98.00 - 6.125 91.875101+00 96.50 - 4.500 92.000102+00 95.00 - 3.125 91.875103+00 93.50 - 2.000 91.500

104+00 92.00 - 1.125 90.875105+00 90.50 - 0.500 90.000106+00 89.00 - 0.125 88.875EVC 107+00 87.50 - 0.000 87.500

Example 13:

A 600 ft VC connects a +4% grade to a -2% grade at station 25+60.55 and elevation 648.64 ft.

Calculate the location and elevation of BVC, the high point, EVC, and the curve elevation at

stations 24+00 and 27+00 ?


PVI

Sta 25+60.55

Elev 648.64

BVC

Sta 22+60.55

EVC

Sta 28+60.55

+ 4% - 2% E

High Point

300' 300'

XHP

Sta. of BVC = Sta. of PVI -2

L = 2560.55 –

600

2

= 2260.55 = (22+60.55)

Elev. of BVC = Elev. of PVI -G1 L

100 2

= 648.64 -4 600

100 2

= 648.64 - 12 = 636.64'








- 45 -

AL (G2-G1)L (-2-4)(600) (-6)(600)E= = = = =-4.5'

800 800 800 800

Elevation for Station 24+00

( ) 2G2-G1G1Elev. of Sta. 24 + 00 = Elev. of BVC + +

100 200L

x x

where x = Sta. 24+00 - Sta. BVC = 2400 - 2260.55 = 139.45′

24 ( 2 4)Elev. Sta. 24 00 636.64 (139.45) 139.45 641.25'

100 (200)(600)

− − + = + + =

or2

G1Elev. of Sta. 24 00 Elev. of BVC 4E

100 L

x x

+ = + +

24 139.45

Elev. of Sta. 24 00 636.64 (139.45) 4( 4.5)100 600

+ = + + −

= 641.25

The High Point : '4006

4*600

)2(4

4*600

2G1G

1LGHPX ==−−=−=

Point x (ft) Tangent Elevation (ft)(This is the tangent that passes through BVC)

VC Offset*(ft)

2

y 4EL

′ =

x

VC Elevation (ft)

BVC 22+60.55 0 636.64 0.00 636.6424+00.00 139.45 642.22 - 0.97 641.25

HP 26+60.55 400 652.64 - 8.00 644.6427+00.00 439.45 654.22 - 9.66 644.56

EVC 28+60.55 600 660.64 - 18.00 642.64* Note that the VC Offset was calculated with respect to gradient G1 only in this table.

An alternative method is to calculate the “VC Offset” with respect to gradients +G1 and G2, sowe can get the following table instead:Point x (ft) x* (ft) Tangent

Elevation (ft)(This tangent iseither +G1 or

G2)

VC Offset (ft)2

y 4EL

′ =

x

VC Elevation (ft)

BVC 22+60.55 0 0 636.64 0.00 636.64

24+00.00 139.45 139.45 642.22 - 0.97 641.25HP 26+60.55 400 200 646.64 - 2.00* 644.6427+00.00 439.45 160.55 645.85 - 1.29* 644.56

EVC 28+60.55 600 0 642.64 0.00* 642.64*It is important to note that x* is measured from right to left (i.e., from EVC to left). The term x* is used instead of x in the “VC Offset” formula above only when x is larger than L/2, that is300 ft in this example. Hence, in this example, x*= 600- x for x larger than L/2 and x*= x when x is smaller than L/2.








- 46 -

4. SIGHT DISTANCE

4.1 Types of Sight Distance

Sight distance is defined as the distance at which objects become visible to the driver. The

design concept of Sight Distance is that drivers traveling at design speed should be able to stop

their vehicles (Stopping Sight Distance SSD) or overtake slower vehicles safely (Passing Sight

Distance PSD).

Minimum SSD

The minimum distance sufficient to enable a driver traveling at design speed to stop his/her

vehicle before striking an unexpected object on the highway.

Minimum SSD = 1.47 V t +)gf (30

V2

±

Where,

SSD is in ft.

V = Design speed, mph

t = perception and reaction time. ( ~ 2.5 seconds)

f = friction factor, depends on tire, surface and speed

g = grade as decimal not as percent (e.g., 6%, g = 0.06). +ve for upgrade and -ve for

downgrade.

+ -








- 47 -

SSD is provided in tables for different pavement conditions

(AASHTO 2001, Tables III-1 and III-2).

Table 16.2. Coefficients of Skidding Friction

Reference: See page 16-5 of the Civil Engineering

Reference Manual, 6th

Edition, by Michael Lindeburg

Exhibit 3-1 and Exhibit 3-7 of AASHTO 2001

Minimum Sight Distances

AASHTO 2001, pages 112 and 124.BC: bituminous concrete, dry

SA: sand asphalt, dry

RA: rock asphalt, dry

CC: portland cement concrete, dry

wet: all wet pavements

stopping sight distance

wet pavements

passing sight

distances

design

speed

mph

Calculatedsight

distance

ft

Designstopping

sightdistance

ft

Assumed passingspeed,mph

Distance2-lane

highway,ft

condition BC SA RA CC wet

new tires

11 mph

20

30

40

5060

70

.74

.76

.79

.75

.75

.75

.79

.75

.78

.76

.74

.74

.76

.73

.78

.76

.40

.36

.33

.31

.30

.29

20

30

40

50

60

70

111.9

196.7

300.6

423.8

566.0

727.6

115

200

305

425

570

730

28

36

44

51

57

64

710

1090

1470

1835

2135

24801 t p is also known as the PIEV time. This name is an acronym for the various

elements of reaction time, including perception, identification, emotion, andvolition

badly worn tires

11 mph20

30

40

50

60

70

.61

.60

.57

.48

.66

.57

.48

.39

.73

.65

.59

.50

.68

.50

.47

.33

.40

.36

.33

.31

.30

.29

2 For the purpose of determining minimum stopping sight distances, t p is takenas 2.5 seconds. For determining passing sight distances, t p is taken as 3.5 to 4.5seconds3 It is assumed that the driver’s eyes are 3.5 feet above the surface of the

roadway. The object being viewed (e.g., an oncoming car) is assumed to be at aheight of 3.5 feet.

Minimum PSD

The minimum distance sufficient to enable a driver traveling at design speed to pass (overtake)

slower vehicles. Passing sight distance is calculated only in case of two lanes of traffic moving

in opposite directions.








- 48 -

4.2 Sight Distance on Horizontal Curves

For S < L

200

*cos1

DS

M R

−=

)(cos200 1

R

M R

DS

−= − , this is also the same equation in figure

Where;

L = Length of horizontal curve, ft.M = distance from centerline of inside lane to edge of obstruction, ftS = stopping distance along the center of the inside lane, ftR = radius of centerline curvature of inside lane, ftD = degree of curvature

Figure III-24(A), p. 220, AASHTO 1994.

Sight Distance (S)Highway

Inside Lane

Line of Sight

M

R

D

Sight

Obstruction

LC LC

V = 1 1 0 k m / h S = 2 4 6 .4 m

10

20

40

60

80

100

200

400

600

800

1000

2000

4000

6000

V = 1 2 0 k m / h S = 2 8 5 .6 m V = 1 0 0 k m / h S = 2 0 5 .0 m

V = 9 0 k m / h S = 1 6 8 .7 m

V = 8 0 k m / h S = 1 3 9 .4 m

V = 7 0 k m / h S = 1 1 0 .8 m

V = 6 0 k m

/ h S = 8 4 . 5 m

V = 5 0 k m / h S = 6 2 . 8 m

V = 4 0 k m

/ h S = 4 4 . 4 m

V = 3 0 0 k

m / h S = 2 9 . 6 m

( M e a s u r e d a l o n g C .L . I n s i d e L a n e ) R a d i u s ,

R ,

C e n t e r l i n e o f I n s i d e L a n e

( m )

Line of sight

S i g h t Dis tance ( S )

H i g h w a y

I n s i d e L

ane

Sightobstruction

Min. R when

a = 10%

Min. R when

a = 10%

28.65 SM =R 1- cos

Rwhere

S=Stopping Sight

Distance

M=Middle OrdinateR=Radius

R

M

Min. R when

a = 10%

18161412108642 Middle Ordinate, MCenterline of Inside Lane to Sight Obstruction (M) in meters








- 49 -

Derivation of Horizontal Sight Distance Formula

M = )]2

cos(1[R )2

cos(*R R oo θ−=θ−

S =180

R o π

θ

θ o =R *

S*180

π=

R

S*30.57

get M = )]

R

S65.28cos(1[R −

S

θθ

θ

Example 14:

The corner of a building is situated next to a horizontal curve with a radius of 132 ft on a rural

highway. The inside lane is 10 ft wide and the inside edge of the road is 6 ft from the corner of a

building. Determine what speed limit should be imposed on this section of the highway.

Assume shoulder width = 4 feet, t =2.5 sec, f = 0.40 (wet conditions), g = 0 (level grade).


R = 132 –10

2

= 127 ft

M = 6 + 4 + 5 = 15 ft

S = 1127 127 15cos 125 ft

28.65 127

− − =

Note:S is calculated using the same equation shown inside AASHTO Figure III-24(A) on page

26.

S = 125 ft = 1.47 V t +)gf (30

V 2

±

125 = 1.47 V (2.5) +)4.0(30

V2

solve for V = 22.5 mph speed limit should be 22.5 mph (~ 20 mph)








- 50 -

Example 15:

A 100 ft right-of-way (R/W) highway with 1250 ft proposed minimum centerline curvature is

being located through the Redwoods. A particularly fine tree stands 80.1 ft from the P-line and

321.3 ft from the PI of tangents which intersect at 135o.1. What is the largest radius curve joining the tangents such that the tree will not encroach on

the right-of-way (R/W)?

2. What is the available stopping-sight distance if the roadway section consists of a 16 ft

median and two 12 ft lanes each way?

See figure below.

P-line








- 51 -


1. O.PC.PI (big triangle)

tan 22.5 =R

3.321X +

X = 0.41421 R - 321.3 => (1)

2. O.C.T (small triangle)

OT2 = CT2 + OC2

(R-50)2 = X2 + (R-80.1)2

R 2 - 100R + 2500 = X2 + R 2 -160.2 R +6416.01

60.2 R - 3916.01 = X2

=> (2) substitute (1) into (2)

60.2 R - 3916.01 = (0.41421 R - 321.3) 2

R = 1480.4 ft

--Note: This is not the R that we use in sight distance equation!








- 52 -

3. Cross section at the tree location

M = 50 - (8+12+6) = 24 ft

28.65M 1 cos

S R

R

= −

R to be used in sight distance formula = 1480.4 - (8+12+6) = 1480.4 - 26 = 1454.4 ft

Note that this is the R that we need to use in the sight distance equation.

28.6524 1454.4 1 cos

1454.4

S = −

S = 529.16 ft

S ~ 161.3 meters

AASHTO Exhibit 3-1 or Table III-3, or Figure III-24(A) on page 48 (after conversion to

proper units). Enter figure with M and R in metric units and with interpolation

See where the red arrows meet on this figure.

S= 161.3 m ⇒ Speed = 90 kph ~ 56.25 mph.








- 53 -

4.3

Sight Distance on Vertical Curves

Design of Vertical Curves Based on Sight Distance

1. CREST:

2. SAG:

Design of Vertical Curves

Crest Sag

S < L

( )2

21

2

hh200

SAL

+=

( )β+=

tanSH200

SAL

2

S > L ( )A

hh200S2L

2

21 +−=

L = 2S- ( )

A

tanSH200 β⋅+

Where S = Sight distance, feet (SSD or PSD)

L = Length of vertical curve, feet

,GGA 12 −= in percent

=1h height of driver's eye, in feet

(used for both SSD and PSD)

=2h For SSD, height of object in feet

For PSD, height of oncoming vehicle

S = Sight distance, feet SSD

L = Length of vertical curve, feet

,GGA 12 −= in percent

H = headlight height.

β = beam angle.

Default

Values

=1h 3.5 ft

=2h 2 ft (SSD) or 3.5 ft (PSD)

H = 2 ft.

β = 1°

L

L

S < L

S > L

β

LS < L

β

LS > L

h2

h1

h1 h2








- 54 -

Minimum length is that which meets all required criteria.

Table III-3 Stopping Sight Distance on Wet Pavements(AASHTO 2001, Exhibit 3-1, p. 112) –See Appendix A for Metric Units.

DesignSpeed(mph)

Brake ReactionDistance

(ft)

BrakingDistanceon Levela

Stopping Sight Distance

Calculated(ft)

Rounded forDesign

(ft)15202530354045505560

65707580

55.173.591.9

110.3128.6147.0165.4183.8202.1220.5

238.9257.3275.6294.0

21.638.460.086.4117.6153.6194.4240.0290.3345.5

405.5470.3539.9614.3

76.7111.9151.9196.7246.2300.6359.8423.8492.4566.0

644.4727.6815.5908.3

80115155200250305360425495570

645730820910

Note: Brake reaction distance predicted on a time of 2.5 s; deceleration rate of 3.4 m/s [11.2 ft/s ] used todetermine calculated sight distance.

Table III-4 Design Controls for Stopping Sight Distance and for Crest Vertical Curves(AASHTO 2001, Exhibit 3-76, p. 274)-- See Appendix A for Metric Units

Designspeed(mph)

Stopping SightDistance

(ft)

Rate of Vertical Curvature, K a

[length (ft) per percent of A]

Calculated Rounded

for Design

15202530354045505560657075

80

80115155200250305360425495570645730820

910

3.06.111.118.529.043.160.183.7

113.5150.6192.8246.9311.6

383.7

37

121929446184114151193247312

384

aRate of vertical curvature, K, is the length of curve per percent algebraic difference in intersecting grades (A).

LK=

A.

Source : A Policy on Geometric Design of Highway and Streets, copyright 2001 by the American Associationof State Highway and Transportation Officials, Washington, DC.








- 55 -

Table III-5 Design Controls for Sag Vertical Curves(AASHTO 2001, Exhibit 3-79, p. 280) --See Appendix A for Metric Units

Designspeed(mph)

Stopping SightDistance

(ft)

Rate of Vertical Curvature, K a


Calculated Rounded

for Design15202530354045505560657075

80

80115155200250305360425495570645730820

910

9.416.525.536.449.063.478.195.7

114.9135.7156.5180.3205.6

231.0

1017263649637896115136157180206

231aRate of vertical curvature, K, is the length of curve per percent algebraic difference in intersecting grades (A).

LK=

A.

Source : A Policy on Geometric Design of Highway and Streets, copyright 2001 by the American Association ofState Highway and Transportation Officials, Washington, DC.

Table III-6 Design Controls for Crest Vertical Curves Based on Passing Sight Distance(AASHTO 2001, Exhibit 3-77, p. 276) --See Appendix A for Metric Units

Design Speed

(mph)

Passing Sight

Distance

(ft)

Rate of Vertical Curvature, K ,a

Rounded for Design


20253035404550556065

707580

710900

10901280147016251835198521352285

248025802680

1802894245857729431203140716281865

219723772565

aRate of vertical curvature, K, is the length of curve per percent algebraic difference in intersecting grades (A).

LK=

A.

Source : A Policy on Geometric Design of Highway and Streets, copyright 2001 by the American Association of StateHighway and Transportation Officials, Washington, DC.








- 56 -

Example 16

A car is travelling at 40 mph up a hill with a +1.25% grade. The descending grade is –2.75%.

What is the required length of curve for proper stopping sight distance?


The design speed (assumed travel speed) = 40 mph

See either Table III-3 or III-4 for SSD ⇒ SSD = 305 ft

Use table of crest vertical curves ⇒ K= 44 ft/percent of A

|A| = |-2.75-1.25| = 4%

Lmin = K|A| = (44) (4) = 176 ft

Check formula :

S = 305 > L = 176

L computed =

2

'200 3.5 2

2(305) 704

+ − = (305 > 70) Initial Assumption is O.K.

Recommended value for Lmin is 176 ft.

Note:

Sight distance tables in AASHTO meet additional criteria in addition to SSD and PSD minimum

length requirement. These tables meet the comfort and drainage criteria, so the length rounded

from these tables can be longer than what the equation calculations show. Unless you are asked

to use the equations for sight distance on vertical curves, it is recommended to use these tables

instead of the equations.








- 57 -

APPENDIX –A-

METRIC TABLES FOR SIGHT DISTANCE ON VERTICAL CURVES








Table 3: Design Controls for Sag Vertical Curves

Design

Speed

(km/hr)

Assumed

Speed for

Condition

(km/hr)

Coefficient of

Friction, f

Stopping

Sight Distance for

Design (m)

Rate of Vertical Curvature, K

[length (m) per percent of A]

Computed Rounded for

Design

30 30-30 0.40 29.6-29.6 3.88-3.88 4-4

40 40-40 0.38 44.4-44.4 7.11-7.11 8-8

50 47-50 0.35 57.4-62.8 10.20-11.54 11-12

60 55-60 0.33 74.3-84.6 14.45-17.12 15-18

70 63-70 0.31 94.1-110.8 19.62-24.08 20-25

80 70-80 0.30 112.8-139.4 24.62-31.86 25-32

90 77-90 0.30 131.2-168.7 29.62-39.95 30-40100 85-100 0.29 157.0-205.0 36.71-50.06 37-51

110 91-110 0.28 179.5-246.4 42.95-61.68 43-62

120 98-120 0.28 202.9-285.6 49.47-72.72 50-73

Table 4: Design Controls for Crest Vertical Curves Based on Passing Sight Distance

Design Speed(km/hr)

Minimum Passing

Sight Distancefor Design

(m)

Rate of Vertical Curvature, K ,a

Rounded for Design

[length (m) per percent of A]

30 217 50

40 285 90

50 345 130

60 407 180

70 482 250

80 541 310

90 605 390

100 670 480




TTE3810 Geometric Design Part One_Fall 2013.pdf

Documents

Transcript of TTE3810 Geometric Design Part One_Fall 2013.pdf