Fundamentals of Transportation/Grade






Filbert Street in San Francisco (North Beach)


Road Vehicle Performance forms the basics that make up highway design guidelines and traffic analysis. Roads need to designed so that the vehicles traveling upon them are accommodated. For example, the Interstate Highway system in the United States has a maximum allowable grade that can be used, so that the semitrucks that frequently use these roads would be able to travel them without encountering grade problems. While this can pose quite a design challenge in mountainous regions like Colorado Rockies, without it those trucks would be forced to find better alternative routes, which costs time and money. The ability of a certain vehicle type to use a road is dependent on the power produced by its motor, as well as other road-based and environmental characteristics. When choosing a road design, these elements need to be considered.




Contents






  • 1 Illustration


  • 2 Fundamental Characteristics


    • 2.1 Aerodynamic Resistance


    • 2.2 Rolling Resistance


    • 2.3 Grade Resistance


    • 2.4 Tractive Effort


    • 2.5 Grade Computation




  • 3 Examples


    • 3.1 Example 1: Racecar Acceleration


    • 3.2 Example 2: Going Up a Hill




  • 4 Thought Question


  • 5 Sample Problem


  • 6 Additional Questions


  • 7 Variables


  • 8 Key Terms





Illustration




Lombard Street in San Francisco


How steep is the grade in San Francisco?
The famous Lombard Street, the curviest street in the world only has a grade of 14.3%. Sources vary on the steepest street in San Francisco. Filbert Street between Hyde and Leavenworth has a grade of 31.5%, Prentiss between Chapman and Powhattan reportedly has a grade of 37%.


Driving in San Francisco at times is like a roller coaster, you can’t see below the nose of your car sitting at the top of the hill, though you can usually see the base of the hill.


What is max grade allowed by law?


Maximum grade is not regulated so much by law as by engineering standards. Maximum grade varies by type of road, and expected speed. In practice, it depends on the alternatives: Is the alternative no road at all? Max grade in the relatively flat Minnesota might be lower than Max grade in mountainous Colorado where there are fewer alternatives. Some typical values for illustration:



  • An interstate is out of standard if it has a grade > 7%.

  • The National Road (built in 1806) had a maximum grade of 8.75%.

  • Local roads are much higher (12% or 15% are sometimes allowed)

  • Otter Tail MN County roads 6%, alleys 8%

  • Driveways can be as much as 30% for a short distance



Fundamental Characteristics




Forces acting on a vehicle


Tractive effort and resistance are the two main forces that oppose one another and determine the performance of roadway vehicles. Tractive effort is the force exerted against the roadway surface to allow a vehicle to move forward. Resistance encompasses all forces that push back and impede motion. Both of these are in units of force. The general formula for this is outlined below:


Ft=ma+Ra+Rrl+Rg{displaystyle F_{t}=ma+R_{a}+R_{rl}+R_{g},!}{displaystyle F_{t}=ma+R_{a}+R_{rl}+R_{g},!}


Where:




  • Ft{displaystyle F_{t},!}{displaystyle F_{t},!} = Tractive Effort


  • m{displaystyle m,!}{displaystyle m,!} = Vehicle Mass


  • a{displaystyle a,!}{displaystyle a,!} = Acceleration


  • Ra{displaystyle R_{a},!}{displaystyle R_{a},!} = Aerodynamic Resistance


  • Rrl{displaystyle R_{rl},!}{displaystyle R_{rl},!} = Rolling Resistance


  • Rg{displaystyle R_{g},!}{displaystyle R_{g},!} = Grade Resistance


These components are discussed in greater detail in the following sections.



Aerodynamic Resistance


Aerodynamic resistance is a force that is produced by turbulent air flow around the vehicle body. This turbulence is dependent on the shape of the vehicle, as well as the friction of air passing over the vehicle's surface. A small portion of this resistance comes from air flow through vehicle components, such as interior ventilation. This resistance can be estimated through the following formula:


Ra=ρ2ACDV2{displaystyle R_{a}={frac {rho }{2}}AC_{D}V^{2},!}{displaystyle R_{a}={frac {rho }{2}}AC_{D}V^{2},!}


Where:




  • ρ{displaystyle rho ,!}{displaystyle rho ,!} = Air Density


  • A{displaystyle A,!}{displaystyle A,!} = Frontal area of the vehicle


  • CD{displaystyle C_{D},!}{displaystyle C_{D},!} = Coefficient of Drag


  • V{displaystyle V,!}{displaystyle V,!} = Speed of the vehicle


Air density is a function of elevation and temperature. Frontal area and coefficient of drag are generally unique to each vehicle or type of vehicle.



Rolling Resistance


Rolling resistance is caused by the interaction of tires with the roadway surface. Three main causes exist that create this resistance. The first is the rigidity of the tire and the roadway surface. The second is tire pressure and temperature. The third is vehicular operating speed. This value of rolling friction can be calculated from a very simplified formula, given here in metric. V{displaystyle V}V is in meters per second.


frl=0.01(1+V44.73){displaystyle f_{rl}=0.01(1+{frac {V}{44.73}}),!}{displaystyle f_{rl}=0.01(1+{frac {V}{44.73}}),!}


The resistance caused by this friction will increase as weight is added to the vehicle. Therefore, rolling resistance can be calculated.


Rrl=frlW{displaystyle R_{rl}=f_{rl}W,!}{displaystyle R_{rl}=f_{rl}W,!}


Where:




  • W{displaystyle W,!}{displaystyle W,!} = Vehicle Weight


  • frl{displaystyle f_{rl},!}{displaystyle f_{rl},!} = Rolling Friction



Grade Resistance


Grade resistance is the simplest form of resistance. It is the gravitational force acting on the vehicle. This force may not be exactly perpendicular to the roadway surface, especially in situations when a grade is present. Thus, grade resistance can be calculated in the following formula:


Rg=WG{displaystyle R_{g}=WG,!}{displaystyle R_{g}=WG,!}


Where:




  • W{displaystyle W,!}{displaystyle W,!} = Vehicle Weight


  • G{displaystyle G,!}{displaystyle G,!} = Grade (length/length)



Tractive Effort


Tractive effort is the force that allows the vehicle to move forward, subject to the resistances of the previous three forces. The derivation of the formula comes from understanding the forces and moments that on around the various tires. It can be summarized into a simple concept, illustrated here.


For a rear-wheel drive car:


Fmax=μW(lf−frlh)/L1−μh/L{displaystyle F_{max}={frac {mu W(l_{f}-f_{rl}h)/L}{1-mu h/L}},!}{displaystyle F_{max}={frac {mu W(l_{f}-f_{rl}h)/L}{1-mu h/L}},!}


For a front-wheel drive car:


Fmax=μW(lr+frlh)/L1+μh/L{displaystyle F_{max}={frac {mu W(l_{r}+f_{rl}h)/L}{1+mu h/L}},!}{displaystyle F_{max}={frac {mu W(l_{r}+f_{rl}h)/L}{1+mu h/L}},!}


Where:




  • Fmax{displaystyle F_{max},!}{displaystyle F_{max},!} = Maximum Tractive Effort


  • μ{displaystyle mu ,!}{displaystyle mu ,!} = Coefficient of road adhesion


  • W{displaystyle W,!}{displaystyle W,!} = Vehicle Weight


  • lr{displaystyle l_{r},!}{displaystyle l_{r},!} = Distance from rear axle to vehicle's center of gravity


  • lf{displaystyle l_{f},!}{displaystyle l_{f},!} = Distance from front axle to vehicle's center of gravity


  • frl{displaystyle f_{rl},!}{displaystyle f_{rl},!} = Coefficient of rolling friction


  • h{displaystyle h,!}{displaystyle h,!} = Height of the center of gravity above the roadway surface


  • L{displaystyle L,!}{displaystyle L,!} = Length of wheelbase



Grade Computation


Most of the work surrounding tractive effort is geared toward determining the allowable grade of a given roadway. With a certain known vehicle type using this road, the grade can be easily calculated. Using the force balance equation for tractive effort, a value for grade can be separating, producing the formula below:


G=Ft−Fa−FrlW{displaystyle G={frac {F_{t}-F_{a}-F_{rl}}{W}},!}{displaystyle G={frac {F_{t}-F_{a}-F_{rl}}{W}},!}


This calculation produces the maximum grade allowed for a given vehicle type. It assumes that the vehicle is operating at optimal engine capacity and, thus, no acceleration can occur, dropping that element from the overall equation.



Examples



Example 1: Racecar Acceleration



TProblem

Problem:

A racecar is speeding down a level straightaway at 100 km/hr. The car has a coefficient of drag of 0.3, a frontal area of 1.5 m2{displaystyle m^{2}}{displaystyle m^{2}}, a weight of 10 kN, a wheelbase of 3 meters, and a center of gravity 0.5 meters above the roadway surface, which is 1 meter behind the front axle. The air density is 1.054 kg/m3{displaystyle m^{3}}{displaystyle m^{3}} and the coefficient of road adhesion is 0.6. What is the maximum possible rate of acceleration for the vehicle?




Example

Solution:

Use the force balancing equation to solve for a{displaystyle a}a.


Ft=ma+Ra+Rrl+Rg{displaystyle F_{t}=ma+R_{a}+R_{rl}+R_{g},!}{displaystyle F_{t}=ma+R_{a}+R_{rl}+R_{g},!}


Since the straightaway is a level one, the grade is zero, thus removing grade resistance from the general problem.


Rg=0{displaystyle R_{g}=0,!}{displaystyle R_{g}=0,!}


Aerodynamic resistance is computed:


Ra=ρ2ACDV2=1.0542(1.5)(0.3)(100∗1000/3600)2=183 N{displaystyle R_{a}={frac {rho }{2}}AC_{D}V^{2}={frac {1.054}{2}}(1.5)(0.3)(100*1000/3600)^{2}=183 N,!}{displaystyle R_{a}={frac {rho }{2}}AC_{D}V^{2}={frac {1.054}{2}}(1.5)(0.3)(100*1000/3600)^{2}=183 N,!}


Rolling resistance is computed:


frl=0.01(1+V44.73)=0.01(1+100∗1000/360044.73)=0.016{displaystyle f_{rl}=0.01(1+{frac {V}{44.73}})=0.01(1+{frac {100*1000/3600}{44.73}})=0.016,!}{displaystyle f_{rl}=0.01(1+{frac {V}{44.73}})=0.01(1+{frac {100*1000/3600}{44.73}})=0.016,!}


Rrl=frlW=0.016(10000)=160 N{displaystyle R_{rl}=f_{rl}W=0.016(10000)=160 N,!}{displaystyle R_{rl}=f_{rl}W=0.016(10000)=160 N,!}


Tractive Effort is computed:


Fmax=μW(lf−frlh)/L1+μh/L=0.6(10000)(1−0.016(0.5))/31−0.6(0.5)/3=2204 N{displaystyle F_{max}={frac {mu W(l_{f}-f_{rl}h)/L}{1+mu h/L}}={frac {0.6(10000)(1-0.016(0.5))/3}{1-0.6(0.5)/3}}=2204 N,!}{displaystyle F_{max}={frac {mu W(l_{f}-f_{rl}h)/L}{1+mu h/L}}={frac {0.6(10000)(1-0.016(0.5))/3}{1-0.6(0.5)/3}}=2204 N,!}


Looking back to the force balancing equation:


ma=Ft−Ra−Rrl−Rg=2204−183−160−0=1861 N{displaystyle ma=F_{t}-R_{a}-R_{rl}-R_{g}=2204-183-160-0=1861 N,!}{displaystyle ma=F_{t}-R_{a}-R_{rl}-R_{g}=2204-183-160-0=1861 N,!}


Divide out mass, which can be computed from weight by dividing out gravity.


W=mg{displaystyle W=mg,!}{displaystyle W=mg,!}


m=Wg=100009.8=1020 kg{displaystyle m={frac {W}{g}}={frac {10000}{9.8}}=1020 kg,!}{displaystyle m={frac {W}{g}}={frac {10000}{9.8}}=1020 kg,!}


Thus, divide mass from the force and acceleration can be found.


a=Fm=18611020=1.82 m/s2{displaystyle a={frac {F}{m}}={frac {1861}{1020}}=1.82 m/s^{2},!}{displaystyle a={frac {F}{m}}={frac {1861}{1020}}=1.82 m/s^{2},!}


Thus, the vehicle is accelerating at a rate of 1.82 meters per second squared.




Example 2: Going Up a Hill



TProblem

Problem:

Using the same case from Example 1, assume that instead the racecar encounters a steep hill that it must travel up. It is desired that the driver maintain the 100 km/h velocity at a very minimum. With that being said, what would be the maximum grade that the hill could be?




Example

Solution:

At the steepest eligible hill, the racecar would be able to maintain 100 km/h without any room for acceleration or deceleration. Therefore, acceleration goes to zero. All other values would stay the same from Example 1. Using the grade formula, the maximum grade can be calculated.


G=Ft−Fa−FrlW=2204−183−16010000=0.1861{displaystyle G={frac {F_{t}-F_{a}-F_{rl}}{W}}={frac {2204-183-160}{10000}}=0.1861,!}{displaystyle G={frac {F_{t}-F_{a}-F_{rl}}{W}}={frac {2204-183-160}{10000}}=0.1861,!}


The maximum allowable grade is 18.61%.




Thought Question


Problem


Why is it that, in mountainous country, trucks and cars have different speed limits?


Answer


Tractive effort is one of the leading reasons, as trucks have a harder time going up steep hills than typical passenger cars, but it is not the only one. Safety is another leading reason, surprisingly, as big rig trucks are obviously more difficult to control in a harsh environment, such as a mountain pass.



Sample Problem


Problem (Solution)



Additional Questions



  • Homework

  • Additional Questions



Variables




  • Ft{displaystyle F_{t}}{displaystyle F_{t}} - Tractive Effort Force


  • Fa{displaystyle F_{a}}{displaystyle F_{a}} - Aerodynamic resistance Force


  • Frl{displaystyle F_{rl}}{displaystyle F_{rl}} - Rolling Resistance Force


  • Fg{displaystyle F_{g}}{displaystyle F_{g}} - Grade Resistance Force


  • W{displaystyle W}W - Vehicle Weight


  • m{displaystyle m}m - Vehicle Mass


  • a{displaystyle a}a - Acceleration


  • ρ{displaystyle rho }rho - Air Density


  • A{displaystyle A}A - Frontal area of the vehicle


  • CD{displaystyle C_{D}}{displaystyle C_{D}} - Coefficient of Drag


  • V{displaystyle V}V - Speed of the vehicle


  • frl{displaystyle f_{rl}}{displaystyle f_{rl}} - Rolling Friction


  • G{displaystyle G}G - Grade


  • μ{displaystyle mu }mu - Coefficient of road adhesion


  • lr{displaystyle l_{r}}{displaystyle l_{r}} - Distance from rear axle to vehicle's center of gravity


  • lf{displaystyle l_{f}}{displaystyle l_{f}} - Distance from front axle to vehicle's center of gravity


  • h{displaystyle h}h - Height of the center of gravity above the roadway surface


  • L{displaystyle L}L - Length of wheelbase



Key Terms



  • Tractive Effort

  • Aerodynamic Resistance

  • Rolling Resistance

  • Grade Resistance




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