Transcript Horizontal Alignment - Center for Transportation Research

Horizontal Alignment
See:
http://www.fhwa.dot.gov/environment/fle
x/ch05.htm (Chapter 5 from FHWA’s
Flexibility in Highway Design)
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Horizontal Alignment
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Design based on appropriate relationship between
design speed and curvature and their interaction
with side friction and superelevation
Along circular path, inertia causes the vehicle to
attempt to continue in a straight line
Superelevation and friction between tire and
roadway provides a force to offset the vehicle’s
inertia; this force is directed toward the center of
curvature (often called centrifugal or centripetal
force)
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Horizontal Alignment
Tangents
2. Curves
3. Transitions
Curves require superelevation (next lecture)
Reason for super: banking of curve, retard
sliding, allow more uniform speed, also
allow use of smaller radii curves (less
land)
1.
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Radius Calculation
Rmin = ___V2______
15(e + f)
Where:
V = velocity (mph)
e = superelevation
f = friction (15 = gravity and unit conversion)
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Radius Calculation
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Rmin related to max. f and max. e allowed
Rmin use 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 changes 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
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Controlled by 4 factors:
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Climate conditions (amount of ice and snow)
Terrain (flat, rolling, mountainous)
Frequency of slow moving vehicles who might be
influenced by high superelevation rates
Highest in common use = 10%, 12% with no ice and
snow on low volume gravel-surfaced roads
8% is logical maximum to minimized slipping by
stopped vehicles
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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%?
Rmin = _____602________________
15(0.04 + 0.12)
Rmin = 1,500 feet
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Curve Types
1.
2.
3.
4.
Simple curves with spirals
Broken Back – two curves same direction
(avoid)
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)
Reverse curves – two curves, opposite direction
(require separation typically for superelevation
attainment)
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Important Components of Simple
Circular Curve
See: ftp://165.206.203.34/design/dmanual/02a-01.pdf
1.
2.
3.
4.
See handout
PC, PI, PT, E, M, and 
L = 2()R()/360
T = R tan (/2)
Source: Iowa DOT
Design Manual
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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])
R
Where:
m = line of sight
S = stopping sight distance
R = radius
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Sight Distance Example
A 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 Example
SSD = 1.47(35 mph)(2.5 sec) +
_____(35 mph)2____ = 246 feet
30(__11.2___  0)
32.2
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Sight Distance Example
m = R(1 – cos [28.65 S])
R
m = 800 (1 – cos [28.65 {246}]) = 9.43 feet
800
(in radians not degrees)
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Horizontal Curve Example
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Deflection angle of a 4º curve is 55º25’, PC at
station 238 + 44.75. 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
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Horizontal Curve Example
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D = 4º
 = 55.417º
R = 1,432.4 ft
L = 2R = 2(1,432.4 ft)(55.417º) = 1385.42ft
360
360
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Horizontal Curve Example
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D = 4º
 = 55.417º
R = 1,432.4 ft
L = 1385.42 ft
T = R tan  = 1,432.4 ft tan (55.417) = 752.30 ft
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2
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Horizontal Curve Example
Stationing goes around horizontal curve.
What is station of PT?
PC = 238 + 44.75
L = 1385.42 ft = 13 + 85.42
Station at PT = (238 + 44.75) + (13 + 85.42) =
252 + 30.17
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Suggested Steps on Horizontal
Design
1. Select tangents, PIs, and general curves make
sure you meet minimum radii
2. 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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