Sight Distance for horizontal curves
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Transcript of Sight Distance for horizontal curves
Sight Distance for Sight Distance for Horizontal CurvesHorizontal Curves
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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
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S
R
M
O
TASight Obstruction on for Horizontal Curves
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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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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
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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
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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’
800
(in radians not degrees)
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SuperelevationSuperelevationDR ABDUL SAMI QURESHIDR ABDUL SAMI QURESHI
ML
N
a
EB
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Motion on Circular Curves
dtdvat
Rvan
2
C.FWeight of VehicleFriction force
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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
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•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
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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
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Theo-retical
Consi-deration
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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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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
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Side Friction Factor
from AASHTO’s A Policy on Geometric Design of Highways and Streets 2004
New Graph
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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.
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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
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Crown / super elevation runoff length
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Source: CalTrans Design Manual online, http://www.dot.ca.gov/hq/oppd/hdm/pdf/chp0200.pdf
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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
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Source: CalTrans Design Manual online, http://www.dot.ca.gov/hq/oppd/hdm/pdf/chp0200.pdf
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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.
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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.
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Attaining Superelevation (1)
•Location of inside edge, centre line, and out side edge are shown relative to elevation of centerline
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Attaining Superelevation (2)
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Attaining Superelevation (3)
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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.
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Superelevation Runoff and Tangent Run out (Crown
Runoff)
Normal cross section
Fully superelevated cross section
Cross section with the adverse crown removed
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Location of Runout and Runoff
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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
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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
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Source: CalTrans Design Manual online, http://www.dot.ca.gov/hq/oppd/hdm/pdf/chp0200.pdf
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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)
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Length of Superelevation Runoff
α = multilane adjustment factorAdjusts for total width
r
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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)
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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
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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
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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.
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Multilane Adjustment factor
• Runout and runoff must be adjusted for multilane rotation.
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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 •
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Lr = 12eα = (12) (0.04) (1.5) G 0.5 Lr = 144 feet
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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
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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.
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Transition curve
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Spiral Curve Spiral Curve TransitionsTransitions
52Source: Iowa DOT Design Manual
SPIRAL TERMINOLOGY
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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
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Tangent-to-Curve Transition
(Appearance)• Improves appearance of the
highways and streets • Provides natural path for drivers
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Tangent-to-Curve Transition (Superelvation or curve
widening requirement)• Accommodates distance needed
to attain super elevation• Accommodates gradual roadway
widening
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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
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Shape of transition curves
• Spiral (Clotoid)= mostly used• Lemniscates (rate of change of
radius not constant)• Cubic parabola
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Transition Curves - Spirals
The Euler spiral (clothoid) is used. The radius at any point of the spiral varies inversely with the distance.
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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)
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• V= 50 mph• C = 3 ft p c. sec• R= 929• Ls = 141 ft
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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)
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Maximum Length of Spiral• Safety problems may occur when
spiral curves are too long – drivers underestimate sharpness of approaching curve (driver expectancy)
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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)
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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.
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Location of Runouts and Runoffs
• Tangent runout proceeds a spiral• Superelevation runoff = Spiral curve
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67Source: Iowa DOT Design Manual
68Source: Iowa DOT Design Manual
69Source: Iowa DOT Design Manual
70Source: Iowa DOT Design Manual
SPIRAL TERMINOLOGY
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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
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With Spirals
Tangent runout (A to B)
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With Spirals
Removal of crown
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With Spirals
Transition of superelevation Full superelevation
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Transition ExampleGiven:• PI @ station 245+74.24• D = 4º (R = 1,432.4 ft) = 55.417º • L = 1385.42 ft
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With no spiral …• T = 752.30 ft• PC = PI – T = 238 +21.94
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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.
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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
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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’
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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
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• 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
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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
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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
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Notes – Iowa DOT
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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
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Quiz Answers5. Increase clear zone6. Improve horizontal and vertical
alignment7. Assure adequate surface drainage8. Increase skid resistance on
downgrade curves
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Some of Your Answers Decrease posted speed Add rumble strips Bigger or better signs Guardrail Better lane markers Sight distance Decrease radius