(IV) VEHICLE STEADY STATE ROAD LOAD REQUIREMENTS

Date: 22nd of September, 2020

Mohd Syaifuddin bin Mohd
Department of Mechanical Engineering
 AT THE END OF THIS LECTURE, STUDENTS WILL UNDERSTAND:
 FORCES, WHEEL TORQUE, WHEEL POWER & ENERGY REQUIRED FOR A
VEHICLE DRIVETRAIN.
 HOW TO CALCULATE STEADY STATE VEHICLE ROAD LOAD REQUIREMENTS.
• A vehicle, consisting of thousands of components, is a complex system.
•This course focuses on vehicle performance, such as speed, gradeability, acceleration,
fuel consumption, and braking performance.
General Description of Vehicle Movement
•The movement behavior of a vehicle along its moving direction is
completely determined by all the forces acting on it in this direction.
•The tractive effort, Ft, in the contact area between the tires of the
driven wheels and the road surface propels the vehicle forward.
•While the vehicle is moving, there is resistance that tries to stop its
movement, Fr
•The resistance usually includes tire rolling resistance, aerodynamic
drag, and uphill gradient resistance.
Identify the
resistance
forces!
https://www.youtube.com/watch?v=w3ZpgPozhfg
• Primarily caused by hysteresis in the tire materials.
• Hysteresis can best be described as the
energy lost, usually in the form of a build up
of heat, when a section of vulcanized rubber
is deformed in a regular, constant manner.

• The more you subject a tire to flexing and

deformation the more heat will build up
within the tire.

Stationary Vehicle
• Due to hysteresis in the deformation of
rubber material, the load at loading is
larger than that at unloading at the same
deformation, z, as shown in Figure 2.3

• In another word, the amount of potential

energy stored as deformation of the
rubber when it is under an external force is
not going be recovered 100% when the
external force is removed.
Forces acting on a tyre vs tyre deformation
during loading and unloading
• When the tire is rolling, as
shown in Figure 2.4a, the
leading half of the contact
area is loading and the
trailing half is unloading.

• Consequently, the hysteresis

causes an asymmetric
distribution of the ground
reaction forces.

• The pressure in the leading half of the contact area is larger than that in the
trailing half, as shown in Figure 2.4a.
• This phenomenon results
in the ground reaction force
shifting forward somewhat.

• This forwardly shifted

ground reaction force, with
the normal load acting on
the wheel center, creates a
moment, which opposes
rolling of the wheel.

• On soft surfaces, the rolling resistance is primarily caused by deformation of

the ground surface as shown in Figure 2.4b.
• The moment produced by
the forward shift of the
resultant ground reaction
force is called rolling
resistant moment:

• To keep the wheel rolling,

a force, F, acting on the
center of the wheel is
required to balance this
rolling resistant moment:
•where rd is the effective radius of the tire and fr = a/rd is called the rolling resistance
coefficient
•the rolling resistant moment can be equivalently replaced by a horizontal force acting
on the wheel center in the opposite movement direction of the wheel.

•where P is the normal load acting on the center of the rolling wheel
•When a vehicle is operated on a slope road, the normal load, P, should be replaced by
the component that is perpendicular to the road surface.

•The rolling resistance coefficient, fr, is a function of tire material, tire structure, tire
temperature, tire inflation pressure, tread geometry, road roughness, road material, and
presence or absence of liquids on the road.
• The values given in Table
2.1 do not take into account
their variations with
Speed.
• Based on experimental
results, many empirical
formulas have been proposed
for calculating the rolling
resistance on a hard surface.

where V is vehicle speed in km/h, and f0 and fs

depend on the inflation pressure of the tire
https://www.youtube.com/watch?v=_NPNiyR5cWo
•A vehicle travelling at a
particular speed in air
encounters a force resisting its
motion, referred to as
aerodynamic drag resulting from
two components: shape drag
and skin friction.
• Actual value of drag force on a
vehicle can be obtained from CFD
simulation, through wind tunnel
test or estimated using coast down
vehicle test.
Shape drag: The forward motion of the vehicle pushes the air in front of it. However, the
air cannot instantaneously move out of the way and its pressure is thus increased,
resulting in high air pressure.
Skin friction: Air close to the skin of the vehicle moves almost at the speed of the vehicle
while air away from the vehicle remains still. In between, air molecules move at a wide
range of speeds. The difference in speed between two air molecules produces a friction
that results in the second component of aerodynamic drag.
•Aerodynamic drag is a function of vehicle speed V, vehicle frontal area, Af ,
shape of the vehicle body, and air density, ρ:

•CD is the aerodynamic drag coefficient that characterizes the shape of the vehicle body
Vw is component of the wind speed on the vehicle moving direction, which has a positive
sign when this component is in the same direction of the moving vehicle and a negative
sign when it is opposite to the vehicle speed.
2001 Honda Insight
Cd = 0.25

GM EV 1
Cd = 0.19

Mercedez Benz E Class

Cd = 0.25
 Can you design and
build one for less than
RM200,000? Target
production of 240 units
per year

https://www.youtube.com/watch?v=WrjjV6nGh0g
 •When a vehicle goes up or down a slope, its weight
produces a component that is always directed in the
downward direction

•This component either opposes the forward motion

(grade climbing) or helps
the forward motion (grade descending).
There are several systems for expressing slope:
1. As an angle of inclination to the horizontal. (This is
the angle α opposite the "rise" side of a triangle with
a right angle between vertical rise and horizontal
run.)

2. As a percentage, the formula for which is which

could also be expressed as the tangent of the angle of
inclination times 100. In the U.S., this percentage "grade"
is the most commonly used unit for communicating
slopes in transportation (streets, roads, highways and rail
tracks), surveying, construction, and civil engineering.
•In vehicle performance analysis, only uphill operation is considered. This grading force
is usually called grading resistance.
Grading resistance, when the road angle is small.
• In some literature, the tire rolling resistance and grading resistance together are called
road resistance, which is expressed as
• Note: Values for wheel torque, wheel power, energy
can be calculated from wheel force:

• Wheel torque required, T = F x R, where R is the wheel

radius

• Power required, P = F x V, where V is the vehicle’s

velocity.

• Energy required, E = P x t, where t is the time traveled.

Can you rewrite these equations
in terms of force and torque?
1. Calculate the tractive effort required (or road load in N) from 0 to
180 km/h at 0% grade and 10% grade for a sedan car with the
following parameters:

Aerodynamic drag coefficient, Cd = 0.4

Rolling resistance, fr = 0.015
Vehicle mass (plus 2 passengers) = 1400 kg
Vehicle frontal area = 2.1 m2

(Note that: ρair = 1.23 kg/m3, g = 9.81 m/s2)

2. If the effective wheel radius = 0.21 m, calculate the wheel

torque requirement (in N.m) and power requirement (in kW) at 0%
grade from 0 to 180 km/h. Calculate tractive power required for
the same velocity range and gradient.
 1 Given:
Froll = M·g·Crr·cos(θ); Faero = 0.5·Cd ·ρ·Af ·V2; Fgrade = M·g·sin(θ); Cd = 0.4,
fr = 0.015,
Total resistance force, Ftotal = Froll + Faero + Fgrade; Wheel torque, Twheel = Ftotal x Rwheel ;
M = 1400 kg,
Wheel power required, Ptotal = Ftotal x V ; Energy = Power x time ;
Af = 2.1 m2,
ρair = 1.23 kg/m3,
g = 9.81 m/s2).

Find:
a. Ftractive vs speed
( 0 – 180 km/h at
0% and 10%
gradient).
b. Twheel and Ptractive
vs speed (0 – 180
km/h at 0%
gradient).
2 Given:
Cd = 0.4,
fr = 0.015,
M = 1400 kg,
Af = 2.1 m2,
ρair = 1.23 kg/m3,
g = 9.81 m/s2).

Find:
a. Ftractive vs speed
( 0 – 180 km/h at
0% and 10%
gradient).
b. Twheel and Ptractive
vs speed (0 – 180
km/h at 0%
gradient).
3 Given:
Cd = 0.4,
fr = 0.015,
M = 1400 kg,
Af = 2.1 m2,
ρair = 1.23 kg/m3,
g = 9.81 m/s2).

Find:
a. Ftractive vs speed
( 0 – 180 km/h at
0% and 10%
gradient).
b. Twheel and Ptractive
vs speed (0 – 180
km/h at 0%
gradient).
4 Given:
Cd = 0.4,
fr = 0.015,
M = 1400 kg,
Af = 2.1 m2,
ρair = 1.23 kg/m3,
g = 9.81 m/s2).

Find:
a. Ftractive vs speed
( 0 – 180 km/h at
0% and 10%
gradient).
b. Twheel and Ptractive
vs speed (0 – 180
km/h at 0%
gradient).
5 Given:
Cd = 0.4,
fr = 0.015,
M = 1400 kg,
Af = 2.1 m2,
ρair = 1.23 kg/m3,
g = 9.81 m/s2).

Find:
a. Ftractive vs speed
( 0 – 180 km/h at
0% and 10%
gradient).
b. Twheel and Ptractive
vs speed (0 – 180
km/h at 0%
gradient).
6 Given:
Cd = 0.4,
fr = 0.015,
M = 1400 kg,
Af = 2.1 m2,
ρair = 1.23 kg/m3,
g = 9.81 m/s2).

Find:
a. Ftractive vs speed
( 0 – 180 km/h at
0% and 10%
gradient).
b. Twheel and Ptractive
vs speed (0 – 180
km/h at 0%
gradient).
7 Given:
Cd = 0.4,
fr = 0.015,
M = 1400 kg,
Af = 2.1 m2,
ρair = 1.23 kg/m3,
g = 9.81 m/s2).

Find:
a. Ftractive vs speed
( 0 – 180 km/h at
0% and 10%
gradient).
b. Twheel and Ptractive
vs speed (0 – 180
km/h at 0%
gradient).
8 Given:
Cd = 0.4,
fr = 0.015,
M = 1400 kg,
Af = 2.1 m2,
ρair = 1.23 kg/m3,
g = 9.81 m/s2).

Find:
a. Ftractive vs speed
( 0 – 180 km/h at
0% and 10%
gradient).
b. Twheel and Ptractive
vs speed (0 – 180
km/h at 0%
gradient).
9 Given:
Cd = 0.4,
fr = 0.015,
M = 1400 kg,
Af = 2.1 m2,
ρair = 1.23 kg/m3,
g = 9.81 m/s2).

Find:
a. Ftractive vs speed
( 0 – 180 km/h at
0% and 10%
gradient).
b. Twheel and Ptractive
vs speed (0 – 180
km/h at 0%
gradient).
10 Given:
Cd = 0.4,
fr = 0.015,
M = 1400 kg,
Af = 2.1 m2,
ρair = 1.23 kg/m3,
g = 9.81 m/s2).

Find:
a. Ftractive vs speed
( 0 – 180 km/h at
0% and 10%
gradient).
b. Twheel and Ptractive
vs speed (0 – 180
km/h at 0%
gradient).
1. Calculate the tractive effort required (or road load in N) from 0 to 250 km/h at 0% grade and 5% grade for
a race car with the following parameters:

Aerodynamic drag coefficient, Cd = 0.3

Rolling resistance, fr = 0.012
Vehicle mass (plus 1 driver) = 800 kg
Vehicle frontal area = 1.6 m2

(Note that: ρair = 1.23 kg/m3, g = 9.81 m/s2)

2. If the effective wheel radius = 0.21 m, calculate the wheel torque requirement (in N.m) and power
requirement (in kW) at 0% grade from 0 to 250 km/h. Calculate tractive power required for the same velocity
range and gradient.

Refer to sample solution on Ulearn