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Lecture 4

# Lecture 4 - The Pennsylvania State University Department of...

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Unformatted text preview: The Pennsylvania State University Department of Civil and Environmental Engineering CE 321: Highway Engineering Lecture 4: Vehicle Performance and Resistances Fall 2007 Overview Vehicle performance Highway design Traffic analysis Resistances encountered by vehicles Aerodynamic Grade Rolling Vehicle Performance Acceleration Freeway merging Entrance ramps Intersection sight distance Deceleration Stopping sight distance Exit ramps Vehicle Performance Ascending and Descending Grades Truck climbing lanes Runaway truck ramps Truck Escape Ramp Truck Lanes Vehicle Performance Highway Design Stopping sight distance Passing sight distance Safety Ramps (Acceleration and deceleration lengths) Maximum grades Sight Distance Vehicle Performance Traffic Analysis Speed limits Signal timing Progression timing (acceleration) Speed limits Vehicle Resistances All forces acting on a vehicle which impede its movement or act against the tractive effort of the vehicle. Expressed in pounds. Tractive effort is the force available, at the roadway surface, to perform work. Depends on available horsepower of vehicle. Expressed in pounds. Forces Acting on a Vehicle Aerodynamic resistance, Ra Rolling resistance, Rrl Grade resistance, Rg Tractive effort of the vehicle, Ff and Fr F = Ff + Fr F = ma + Ra + Rrl + Rg Aerodynamic Resistance Turbulent flow of air over the body of a vehicle (85% of aero resistance) Negligible at low speeds Overwhelming at high speeds (Maglev) Aerodynamic Resistance 2 Ra = CD Af V 2 = air density (slugs/ft3) CD = Coefficient of drag (unitless) Af = Frontal area of vehicle (ft2) V = Velocity of vehicle (ft/s) Aerodynamic Resistance Typical values = air density Altitude (ft) 0 5,000 10,000 Temperature ( F) 59.0 41.2 23.4 Pressure (psia) 14.7 12.2 10.1 Air Density (slugs/ft3) 0.002378 0.002045 0.001755 CD = Drag Coefficient Vehicle Type Automobile Bus Tractor-trailer Motorcycle (with rider) Drag Coefficient 0.25-0.55 0.5-0.7 0.6-1.3 0.27-1.8 Aerodynamic Resistance Aerodynamic Resistance Aerodynamic Resistance Power Required The power required to overcome aerodynamic resistance is significant at high speeds. Since 1 horsepower = 550 ft-lb per second hpRa CD AfV = 1,100 3 Rolling Resistance Resistance resulting from the tire/pavement interface Tire deformation Consider rigidity of tire Hardness of roadway surface Rolling Resistance Rrl = f rlW V f rl = 0.011 + 147 Rrl = rolling resistance (lb.) W = Weight of the vehicle acting normal to the pavement surface. (Wcos) frl = Coefficient of rolling resistance V = Vehicle speed in ft/s Power Required The power required to overcome rolling resistance is not as dependent on speed. Since 1 horsepower = 550 ft-lb per second hpRrl f rlWV = 550 Grade Resistance Gravitational force resisting a vehicles motion Since highway grades are usually small as discussed before, the sin tan . Grade Resistance Rg = W tan g = WG W = Weight of the vehicle acting normal to the pavement surface. G = grade defined as vertical rise per some specified horizontal distance. Percent Grade Illustrated 5% grade = 2.86 100 ft 5 ft Grade Resistance Example A 5,000 lb automobile traveling up a 6% grade. Rg = W tan = WG = 5,000 lbs x 0.06 = 300 lbs A 80,000 lb tractor trailer traveling up a 6% grade. Rg = W tan = WG = 80,000 lbs x 0.06 = 4,800 lbs A 40 car train with cars weighing 120,000 lbs each traveling up a 6% grade. Rg = W tan = WG = 40 x 120,000 lbs x 0.06 = 288,000 lbs Example Problem 1 A 2,000 lb car is traveling at an elevation of 5,000 feet ( = 0.002045 slugs/ft) on a concrete surface. If the car is traveling at 70 mph and has a CD = 0.4 and Af = 20 ft2 and an available tractive effort of 255 lb, what is the maximum grade that this car could ascend and maintain the 70 mph speed? Example 1 (con't) F = Ra + Rrl + Rg Grade Resistance Rg = WG = 2,000G Aerodynamic Resistance 0.002045 2 Ra = CD AfV = (0.4)(20)(70 x1.47)2 2 2 = 86.61 lb Example 1 cont F = Ra + Rrl + Rg Rolling Resistance Rrl = frlW 70 x1.47 = 0.011 + x 2,000 147 = 34 lb Example 1 (con't) F = Ra + Rrl + Rg Therefore: F G = 255 lb = 86.61 + 34 + 2000G = 0.067 or a 6.7% grade Recap Vehicle performance affects all aspects of highway engineering Vehicle resistances are broken into three major categories Aerodynamic Grade Rolling Tractive effort must overcome these forces Available Engine generated ...
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