Horizontal Alignment CE 453 Lecture 16

Objectives Identify curve types and curve components See: http://www.fhwa.dot.gov/environment/flex/ch05.htm (Chapter 5 from FHWA’s Flexibility in Highway Design)

Horizontal Alignment Design based on appropriate relationship between design speed and curvature and their relationship with side friction and superelevation Along circular path, vehicle undergoes centripetal acceleration toward center of curvature (lateral acceleration) Balanced by superelevation and weight of vehicle (friction between tire and roadway)

Horizontal Alignment Tangents Curves 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)

Radius Calculation Rmin = ___V2______ 15(e + f) Where: V = velocity (mph) e = superelevation f = friction (15 = gravity and unit conversion)

Radius Calculation 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)

Max e Controlled by 4 factors: Climate conditions (amount of ice and snow) Terrain (flat, rolling, mountainous) Type of area (rural or urban) Frequency of slow moving vehicles who might be influenced by high superelevation rates

Max e Highest in common use = 10%, 12% with no ice and snow on low volume gravel-surfaced roads 8% is logical maximum to minimize slipping by stopped vehicles For consistency use a single rate within a project or on a highway

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.

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

Radius Calculation (Example) For emax = 4%? (urban situation) Rmin = _____602________________ 15(0.04 + 0.12) Rmin = 1,500 feet

Radius Calculation (Example) For emax = 2%? (rotated crown) Rmin = _____602________________ 15(0.02 + 0.12) Rmin = 1,714 feet

Radius Calculation (Example) For emax = -2%? (normal crown, adverse direction) Rmin = _____602________________ 15(-0.02 + 0.12) Rmin = 2,400 feet

Curve Types Simple curves with spirals (why 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)

Important Components of Simple Circular Curve See: ftp://165.206.254.150/dotmain/design/dmanual/English/e02a-01.pdf   1.     See handout 2.     PC, PI, PT, E, M, and  3.     L = 2()R()/360 4.     T = R tan (/2) Direction of stationing Source: Iowa DOT Design Manual

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

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

Sight Distance Example SSD = 1.47(35 mph)(2.5 sec) + _____(35 mph)2____ = 246 feet 30(__11.2___  0) 32.2

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)

Horizontal Curve Example Deflection angle of a 4º curve is 55º25’, PI at station 245+97.04. 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 4

Horizontal Curve Example D = 4º  = 55.417º R = 1,432.4 ft L = 2R = 2(1,432.4 ft)(55.417º) = 1385.42ft 360 360

Horizontal Curve Example D = 4º  = 55.417º R = 1,432.4 ft L = 1385.42 ft T = R tan  = 1,432.4 ft tan (55.417) = 752.29 ft 2 2

Stationing Example Stationing goes around horizontal curve. For previous example, what is station of PT? First calculate the station of the PC: PI = 245+97.04 PC = PI – T PC = 245+97.04 – 752.29 = 238+44.75

Stationing Example (cont) PC = 238+44.75 L = 1385.42 ft Station at PT = PC + L PT = 238+44.75 + 1385.42 = 252+30.17

Suggested Steps in Horizontal Design Select tangents, PIs, and general curves making sure you meet minimum radius criteria Select specific curve radii/spiral and calculate important points (see lab) using formula or table (those needed for design, plans, and lab requirements) Station alignment (as curves are encountered) Determine super and runoff for curves and put in table (see next lecture for def.) Add information to plans