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.

IllustrationEdit

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 %. The steepest street in San Francisco is Filbert Street between Hyde and Leavenworth at a grade of 31.5 %. reference

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 CharacteristicsEdit

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:

  F_t  = ma + R_a  + R_{rl}  + R_g  \,\!

Where:

  • F_t\,\! = Tractive Effort
  • m\,\! = Vehicle Mass
  • a\,\! = Acceleration
  • R_a\,\! = Aerodynamic Resistance
  • R_{rl}\,\! = Rolling Resistance
  • R_g\,\! = Grade Resistance

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

Aerodynamic ResistanceEdit

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:

R_a  = \frac{\rho }{2}AC_D V ^2 \,\!

Where:

  • \rho\,\! = Air Density
  • A\,\! = Frontal area of the vehicle
  • C_D\,\! = Coefficient of Drag
  • 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 ResistanceEdit

Rolling resistance is caused by a vehicle's internal mechanical friction and 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 is in meters per second.

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.

R_{rl} = f_{rl}W\,\!

Where:

  • W\,\! = Vehicle Weight
  • f_{rl}\,\! = Rolling Friction

Grade ResistanceEdit

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:

R_g = WG\,\!

Where:

  • W\,\! = Vehicle Weight
  • G\,\! = Grade (length/length)

Tractive EffortEdit

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:

F_{max} = \frac{{\mu W (l_f-f_{rl}h)/L}}{{1-\mu h /L}}\,\!

For a front-wheel drive car:

F_{max} = \frac{{\mu W (l_r+f_{rl}h)/L}}{{1+\mu h /L}}\,\!

Where:

  • F_{max}\,\! = Maximum Tractive Effort
  • \mu\,\! = Coefficient of road adhesion
  • W\,\! = Vehicle Weight
  • l_r\,\! = Distance from rear axle to vehicle's center of gravity
  • l_f\,\! = Distance from front axle to vehicle's center of gravity
  • f_{rl}\,\! = Coefficient of rolling friction
  • h\,\! = Height of the center of gravity above the roadway surface
  • L\,\! = Length of wheelbase

Grade ComputationEdit

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 = \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.

ExamplesEdit

Example 1: Racecar AccelerationEdit

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 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/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.

  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.

  R_g = 0  \,\!

Aerodynamic resistance is computed:

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:

f_{rl} = 0.01(1+\frac{{V}}{{44.73}}) = 0.01(1+\frac{{100 * 1000/3600}}{{44.73}}) = 0.016 \,\!

R_{rl} = f_{rl}W = 0.016(10000) = 160\ N \,\!

Tractive Effort is computed:

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 = 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\,\!

m = \frac{{W}}{{g}} = \frac{{10000}}{{9.8}} = 1020\ kg\,\!

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

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 HillEdit

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/hr 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/hr 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 = \frac{{F_t  - F_a  - F_{rl} }}{W} = \frac{{2204  - 183  - 160 }}{10000} = 0.1861
\,\!

The maximum allowable grade is 18.61%.

Thought QuestionEdit

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 ProblemEdit

Problem (Solution)

Additional QuestionsEdit

VariablesEdit

  • F_t - Tractive Effort Force
  • F_a - Aerodynamic resistance Force
  • F_{rl} - Rolling Resistance Force
  • F_g - Grade Resistance Force
  • W - Vehicle Weight
  • m - Vehicle Mass
  • a - Acceleration
  • \rho - Air Density
  • A - Frontal area of the vehicle
  • C_D - Coefficient of Drag
  • V - Speed of the vehicle
  • f_{rl} - Rolling Friction
  • G - Grade
  • \mu - Coefficient of road adhesion
  • l_r - Distance from rear axle to vehicle's center of gravity
  • l_f - Distance from front axle to vehicle's center of gravity
  • h - Height of the center of gravity above the roadway surface
  • L - Length of wheelbase

Key TermsEdit

  • Tractive Effort
  • Aerodynamic Resistance
  • Rolling Resistance
  • Grade Resistance
Last modified on 4 March 2014, at 20:27