sight distance for horizontal curves

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Sight Distance for Sight Distance for Horizontal CurvesHorizontal Curves

1

Provided Sight Distance

• Potential sight obstructions– On horizontal curves: barriers,

bridge-approach fill slopes, trees, back slopes of cut sections

– On vertical curves: road surface at some point on a crest vertical curve, range of head lights on a sag curve

3

S

R

M

O

TASight Obstruction on for Horizontal Curves

4

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

5

Middle ordinate• 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

6

Middle ordinate

• Angle subtended at centre of circle by arc AT is 2θ in degree then

• S / πR = 2θ / 180• S = 2R θπ / 180• θ = S 180 / 2R π = 28.65 (S) / R• R-M/R = cos θ • M = [1 – cos 28.65 (S) / R ]

θR

MA TB

T

O

θ

7

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

8

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

32.2

9

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

800

(in radians not degrees)

10

SuperelevationSuperelevationDR ABDUL SAMI QURESHIDR ABDUL SAMI QURESHI

ML

N

a

EB

11

Motion on Circular Curves

dtdvat

Rvan

2

C.FWeight of VehicleFriction force

12

Definition• The transverse slope provided by

raising outer edge w.r.t. inner edge

• To counteract the effects of C. Force (overturning/skid laterally)

L

N

a

EB

M

13

•S.E. expressed in ratio of height of S.E. expressed in ratio of height of outer edge to the horizontal widthouter edge to the horizontal width

e = NL / ML = tan = NL / ML = tan θtan tan θ = sin = sin θ, , θ is very small is very small

e =NL / MN = E / Be =NL / MN = E / BE = Total rise in outer edgeE = Total rise in outer edgeB total width of pavementB total width of pavement

ML

N

θ

EB

14

FgFW fp

α

α

C cos θ Cx

Mx =M sin θ

My =M cos a

F f

F f

θ

C

M 1 fte

≈Rv

1. C.F2. Weight of Vehicle3. Friction force

X-X

Y-Y

C

N

N

C sin θ

15

Theo-retical

Consi-deration

16

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

17

Selection of e and fs• Practical limits on super elevation (e)

– Climate– Constructability– Adjacent land use

• Side friction factor (fs) variations– Vehicle speed– Pavement texture– Tire condition

• The maximum side friction factor is the point at which the tires begin to skid

• Design values of fs are chosen somewhat below this maximum value so there is a margin of safety

18

19

Side Friction Factor

from AASHTO’s A Policy on Geometric Design of Highways and Streets 2004

New Graph

20

Maximum Superelevation • Superelevation cannot be too large

since an excessive mass component may push slowly moving vehicles down the cross slope.

• Limiting values emax – 12 % for regions with no snow and ice

conditions (higher values not allowed),– 10 % recommended value for regions

without snow and ice conditions,– 8% for rural roads and high speed urban

roads,– 4, 6% for urban and suburban areas.

21

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

highway. The design speed is 70 mph. Using AASHTO standards, what is the maximum super elevation rate for existing curve radius of 2500 ft and 300 ft for safe vehicle operation?

• Assume the maximum super elevation rate for the given region is 8%.

• max e = ?• For 70 mph, f = 0.10• 1. 2500 = V2/15(fs+e) = (70 )2/(0.10 + e) = 0.0306• e = 3%• 2. 300 = V2/(fs+e) = (70 x 1.47)2/32.2(0.10 + e) = 0.988• e = 9.8%

• 300 = V2/g(fs+e) = (70)2/15(fs + 0.8)• f = 1.008 > 0.10

• 300 = V2/15(fs+e) = (V )2/15(0.10+ 0.8)• V = 28.46 mph

22

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Crown / super elevation runoff length

24

Source: CalTrans Design Manual online, http://www.dot.ca.gov/hq/oppd/hdm/pdf/chp0200.pdf

25

Attainment of Superelevation - General

1. Tangent to superelevation 2. Must be done gradually over a distance

without appreciable reduction in speed or safety and with comfort

3. Change in pavement slope should be consistent over a distance

4. Methods (Exhibit 3-37 p. 186) a. Rotate pavement about centerline b. Rotate about inner edge of pavement c. Rotate about outside edge of pavement

26

Source: CalTrans Design Manual online, http://www.dot.ca.gov/hq/oppd/hdm/pdf/chp0200.pdf

27

Common methods of developing the transition to super elevation• At (2)the out side edge is far below the centre

line as the inside edge• At (3)the out side edge has reached the level of

the centre line• At point (4) the out side edge is located as far

above as the inside edge is below the centerline.• Finally , at point (5) the cross section is fully

super elevated and remain through out the circular curve

• The reverse of these profiles is found at the other end of circular curve.

28

Common methods of developing the transition to super elevation• Location of inside edge, centre line, and out side edge are

shown relative to elevation of centerline• The difference in elevation being equal to the normal

crown times the pavement width.• At A the out side edge is far below the centre line as the

inside edge• At B the out side edge has reached the level of the centre

line• At point C the out side edge is located as far above as the

inside edge is below the centerline.• Finally , at point E the cross section is fully super elevated• The reverse of these profiles is found at the other end of

circular curve.

29

Attaining Superelevation (1)

•Location of inside edge, centre line, and out side edge are shown relative to elevation of centerline

30

Attaining Superelevation (2)

31

Attaining Superelevation (3)

32

Superelevation Transition Section

• Tangent Runout (Crown Runoff) Section + Superelevation Runoff Section.

• Tangent runout = the length of highway needed to change the normal cross section to the cross section with the adverse crown removed.

Super elevation runoff• Super elevation runoff = the length of

highway needed to change the cross section with the adverse crown removed to the cross section fully super elevated.

33

34

Superelevation Runoff and Tangent Run out (Crown

Runoff)

Normal cross section

Fully superelevated cross section

Cross section with the adverse crown removed

35

Location of Runout and Runoff

36

Tangent Runout Section

• Length of roadway needed to accomplish a change in outside-lane cross slope from normal cross slope rate to zero

For rotation about centerline

37

Superelevation Runoff Section

• Length of roadway needed to accomplish a change in outside-lane cross slope from 0 to full superelevation or vice versa

• For undivided highways with cross-section rotated about centerline

38

Source: CalTrans Design Manual online, http://www.dot.ca.gov/hq/oppd/hdm/pdf/chp0200.pdf

39

Minimum Length of Tangent RunoutLt = eNC x Lr

ed

where• eNC = normal cross slope rate (%)• ed = design superelevation rate• Lr = minimum length of superelevation runoff (ft)(Result is the edge slope is same as for Runoff

segment)

40

Length of Superelevation Runoff

α = multilane adjustment factorAdjusts for total width

r

41

Minimum Length of Runofffor curve

• Lr based on drainage and aesthetics and design speed.

• Relative gradient is the rate of transition of edge line from NC to full superelevation traditionally taken at 0.5% ( 1 foot rise per 200 feet along the road)

42

Design Requirements for Runoffs

Maximum Relative Gradient

Relative gradient is the rate of transition of edge line from NC to full super elevation

Relative gradient

43

Relative Gradient (G)• Maximum longitudinal slope• Depends on design speed, higher

speed = gentler slope. For example:• For 15 mph, G = 0.78%• For 80 mph, G = 0.35%• See table, next page

44

Maximum Relative Gradient (G)

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.

45

Multilane Adjustment factor

• Runout and runoff must be adjusted for multilane rotation.

46

Length of Superelevation Runoff Example

For a 4-lane divided highway with cross-section rotated about centerline, design superelevation rate = 4%. Design speed is 50 mph. What is the minimum length of superelevation runoff (ft)

Lr = 12eα G •

47

Lr = 12eα = (12) (0.04) (1.5) G 0.5 Lr = 144 feet

48

Tangent runout length

Example continued• Lt = (eNC / ed ) x Lr as defined previously, if NC = 2%Tangent runout for the example is:

LT = 2% / 4% * 144’ = 72 feet

49

From previous example, speed = 50 mph, e = 4%From chart runoff = 144 feet, same as from

calculation

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.

50

Transition curve

51

Spiral Curve Spiral Curve TransitionsTransitions

52Source: Iowa DOT Design Manual

SPIRAL TERMINOLOGY

53

Transition Curves – Spirals (Safety)

• Provided between tangents and circular curves or between two circular curves

• It provides the path where radial force gradually increased or decreased while entering or leaving the circular curves

54

Tangent-to-Curve Transition

(Appearance)• Improves appearance of the

highways and streets • Provides natural path for drivers

55

Tangent-to-Curve Transition (Superelvation or curve

widening requirement)• Accommodates distance needed

to attain super elevation• Accommodates gradual roadway

widening

56

Ideal shape of transition curve• When rate of introduction of C.F. is

consistent• When rate of change C.

Acceleration is consistent• When radius of transition curve

consistently change from infinity to radius of circular curve

57

Shape of transition curves

• Spiral (Clotoid)= mostly used• Lemniscates (rate of change of

radius not constant)• Cubic parabola

58

Transition Curves - Spirals

The Euler spiral (clothoid) is used. The radius at any point of the spiral varies inversely with the distance.

59

Minimum Length of Spiral

Possible Equations: When consistent C.F is considered

Larger of (1) L = 3.15 V3

RCWhere:

L = minimum length of spiral (ft)V = speed (mph)R = curve radius (ft)C = rate of increase in centripetal acceleration (ft/s3) use 1-3 ft/s3 for highway)

60

• V= 50 mph• C = 3 ft p c. sec• R= 929• Ls = 141 ft

61

Minimum Length of Spiral

When appearance of the highways is considered

1.Minimum- 2.Maximum Length of Spiral

Or L = (24pminR)1/2

Where:

L = minimum length of spiral (ft) = 121.1 ftR = curve radius (ft) = 930pmin = minimum lateral offset between the tangent and circular curve (0.66 feet)

62

Maximum Length of Spiral• Safety problems may occur when

spiral curves are too long – drivers underestimate sharpness of approaching curve (driver expectancy)

63

Maximum Length of Spiral

L = (24pmaxR)1/2

Where:

L = maximum length of spiral (ft) = 271R = curve radius (ft)pmax = maximum lateral offset between the tangent and circular curve (3.3 feet)

64

Length of Spiralo AASHTO also provides recommended

spiral lengths based on driver behavior rather than a specific equation.

o Super elevation runoff length is set equal to the spiral curve length when spirals are used.

o Design Note: For construction purposes, round your designs to a reasonable values; e.g.

Ls = 141 feet, round it toLs = 150 feet.

65

Location of Runouts and Runoffs

• Tangent runout proceeds a spiral• Superelevation runoff = Spiral curve

66

67Source: Iowa DOT Design Manual

68Source: Iowa DOT Design Manual

69Source: Iowa DOT Design Manual

70Source: Iowa DOT Design Manual

SPIRAL TERMINOLOGY

71

Attainment of superelevation on spiral

curvesSee sketches that follow:Normal Crown (DOT – pt A) 1. Tangent Runout (sometimes known as crown

runoff): removal of adverse crown (DOT – A to B) B = TS

2. Point of reversal of crown (DOT – C) note A to B = B to C

3. Length of Runoff: length from adverse crown removed to full superelevated (DOT – B to D), D = SC

4. Fully superelevate remainder of curve and then reverse the process at the CS.

72Source: Iowa DOT Standard Road Plans RP-2

With Spirals

Same as point E of GB

73

With Spirals

Tangent runout (A to B)

74

With Spirals

Removal of crown

75

With Spirals

Transition of superelevation Full superelevation

76

77

Transition ExampleGiven:• PI @ station 245+74.24• D = 4º (R = 1,432.4 ft) = 55.417º • L = 1385.42 ft

78

With no spiral …• T = 752.30 ft• PC = PI – T = 238 +21.94

79

For: • Design Speed = 50 mph • superelevation = 0.04 • normal crown = 0.02

Runoff length was found to be 144’Tangent runout length = 0.02/ 0.04 * 144 = 72 ft.

80

Where to start transition for superelevation?

Using 2/3 of Lr on tangent, 1/3 on curve for superelevation runoff:

Distance before PC = Lt + 2/3 Lr =72 +2/3 (144) = 168 Start removing crown at: PC station – 168’ = 238+21.94 - 168.00 = Station = 236+ 53.94

81

Location Example – with spiral

• Speed, e and NC as before and = 55.417º • PI @ Station 245+74.24• R = 1,432.4’• Lr was 144’, so set Ls = 150’

82

Location Example – with spiral

See Iowa DOT design manual for more equations:

http://www.dot.state.ia.us/design/00_toc.htm#Chapter_2

• Spiral angle Θs = Ls * D /200 = 3 degrees• P = 0.65 (calculated)• Ts = (R + p ) tan (delta /2) + k = 827.63 ft

83

• TS station = PI – Ts = 245+74.24 – 8 + 27.63 = 237+46.61Runoff length = length of spiral Tangent runout length = Lt = (eNC / ed ) x Lr = 2% / 4% * 150’ = 75’Therefore: Transition from Normal crown

begins at (237+46.61) – (0+75.00) = 236+71.61

Location Example – with spiral

84

With spirals, the central angle for the circular curve is reduced by 2 * Θs

Lc = ((delta – 2 * Θs) / D) * 100 Lc = (55.417-2*3)/4)*100 = 1235.42 ft Total length of curves = Lc +2 * Ls =

1535.42 Verify that this is exactly 1 spiral length

longer than when spirals are not used (extra credit for who can tell me why, provide a one-page memo by Monday)

Location Example – with spiral

85

Also note that the tangent length with a spiral should be longer than the non-spiraled curve by approximately ½ of the spiral length used. (good check – but why???)

Location Example – with spiral

86

Notes – Iowa DOT

87

Quiz AnswersWhat can be done to improve the safety

of a horizontal curve?

Make it less sharp Widen lanes and shoulders on curve Add spiral transitions Increase superelevation

88

Quiz Answers5. Increase clear zone6. Improve horizontal and vertical

alignment7. Assure adequate surface drainage8. Increase skid resistance on

downgrade curves

89

Some of Your Answers Decrease posted speed Add rumble strips Bigger or better signs Guardrail Better lane markers Sight distance Decrease radius

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