Chapter-2
Acceleration and Braking Performance
•
Acceleration Performance:
–
–
•
Power Limited Acceleration
Traction Limited Acceleration
Braking Performance:
–
–
–
–
–
Braking forces
Braking Performance
Wheel Lockup
Brake Proportioning
ABS
Acceleration Performance
• Acceleration Performance Can be studied as
– Engine Power Limited Acceleration
– Traction Limited Acceleration
 Engine Power Limited Acceleration
– The acceleration performance depends on
• Engine Characteristics
• Power to Weight Ratio
• Power Train Inertia
 Traction Limited Acceleration
– Assuming adequate power from the engine, the acceleration is
limited by the coefficient of performance between the tyre and
road
 Acceleration Performance depends on
• Engine Power is Limiting Factor
• Traction is Limiting Factor
@ Higher Speeds
@ Lower Speeds
 Engine Torque Depends on Mean Effective Pressure
i.e. Volumetric Efficiency
the ratio of the mass of air inducted into the engine cylinder during the suction stroke to the
mass of the air corresponding to the swept volume of the engine at atmospheric pressure and
temperature
@ Higher Speed---- Sucking ability comes down
So Torque comes Down
E
T
F.D
Ng
Nf
Pe, Te
Pe = Engine Power
Pe  2NT / 60000
Te = Engine Torque
NT  Pe * 60000 / 2
Ng = Transmission Gear ratio
Nf = Final Drive gear ratio
N= Speed
T= Torque
Taking Pe = constant
NT  C
T C/N
Torque is Inversely Proportional to Speed
Engine Power Limited Acceleration
 Improve Acceleration performance with out disturbing Engine
E
Pe, Te
T
F.D
Ng
Nf
R.R
• Improve Gear Ratios
• Improve Final Drive
• Reduce mass of Power Train
• Reduce Rolling Resistance
• Reduce Aerodynamic Resistance
Tune Transmission to improve Acceleration Performance
Comments on Engine Characteristics
• For vehicular applications, the ideal performance characteristics of a
power plant are constant output power over the full range
• The engine output torque varies with speed hyperbolically
Power
Torque
Engine Speed
• This characteristic will provide the vehicle with high tractive effort at
low speeds where demands for acceleration, drawbar pull or grade
climbing capability are high
• The series wound DC motor and steam engine provide such
characteristics
• The internal combustion engine has less favorable performance
characteristics, and can be used only with a suitable transmission
Engine Characteristics
Engine characteristic curves.
•
For an ICE in a typical passenger car:- Max T 100-300 Nm (trucks 10 times higher)
•
Max T occurs at 2500-4500 rpm (trucks 1000-1500 rpm)
•
Diesel Engine suitable for Heavier Vehicles
Transmission- Need
Pe = Engine power
Te = Engine Torque
Ne = Engine Speed
Tt = Transmission Torque
Nt = Transmission Speed
Power
Pe, Te, Ne
Transmission
Pe, Tt, Nt
Torque
Speed
Manual and Automatic Transmission
Manual and Automatic Transmission Systems.
Constant Power and Tractive –Effort / RoadSpeed Curves
P  2NT  2NFr  r.F  V .F
P
F  Tractive Effort 
V
TE 
Power at roadwheels Pr

N
Velocity
V
• Power at the roadwheel is the product of tractive effort and
velocity it follows that tractive effort is proportional to the
power and inversely proportional to the road speed.
• A curve of constant power will produce an ideal form of
tractive effort curve.
Curve of Constant Power
• As road speed falls the tractive
effort increases in proportion to
the increased resistance, thereby
providing very stable
conditions.
Manual Transmission
Honda CR-V
Vehicle Honda CRV
1. Gear Ratios
Ng1: 2.786:1
Ng2: 1.614:1
Ng3: 1.082:1
Ng4: 0.773:1
Ng5: 0.566:1
Reverse: 2.000
Final Drive: 4.500
2. Max. Power
180 hp @6800 rpm= 132.5 kW@ 6800 rpm
3. Max. Torque
161 ft-lb@4400 rpm= 118 Nm @4400 rpm
4. Wheel Diameter
17” 225/65 Wheel Dia: 43.18 cm+14.6 cm
=57.78 cm
Determination of Gear Ratios
Tractive Resistance = R.R+A.R+G.R
Rolling Resistance, Aerodynamic Resistance, Gradient Resistance
Tractive Re sis tan ce  ( f rW  1 / 2 V 2CD A  W sin  )
FR  f rW  W sin   W ( f r  sin  ) ignore air resis tan ce
@ Low speed
Equating Tractive Resistance and Tractive Effort
Tractive effort  Te * N tg1 * N fd *tr / rtyre
Te * N tg1 * N fd *tr / rtyre  W ( f r  sin  )
N tg1 
W ( f r  sin  max ) * rtyre
Te * N fd *tr
This equation gives first gear ratio
Final and intermediate Gear Ratio
Vmax 
N tg 4 
 * Dtyre * ( N e rpmmax /( N tg 4 * N fd ) * 60
1000
 *D tyre (m) * N e rpmmax * 60
Vmax (kmph) * N fd *1000
N tg 2
N tg1

kg  3
N tg 3
N tg 2

N tg 4
N tg 3
 kg
N tg 4
N tg1
Using this equation final drive gear ratio can be fixed assuming top gear
ratio as 1
Velocity Calculations
 * Dtyre (m) * (
V  vehicle velocity 
N enginerpm
N tgr * N fd
1000
) * 60
kmph
Calculate Maximum Velocity for each of the gear ratio
Tractive Force and Acceleration Calculations
Fx 
ax 
Te N tf  tf
r
Te N tf  tf
rm
Gear Ratios: Wide and Close
• Wide and Close Gear Ratios:
First Gear
Second Gear
Transmission Gear Ratio
1
4:1
2:1
4/2
=2
200%
Wider
2
4:1
3:1
4/3
=1.3
133%
Closer
• Close-ratio transmissions are generally offered in sports cars, in
which the engine is tuned for maximum power in a narrow range
of operating speeds and the driver can be expected to enjoy
shifting often to keep the engine in its power band
• Narrowing the gaps will increase acceleration at speed, and
potentially improve top speed under certain conditions, but
acceleration from stopped and operation in traffic will suffer.
Gear Ratios
McLaren F1 LM
3.23:1, 2.19:1, 1.71:1,
1.39:1, 1.16:1, 0.93:1,
with a final drive of 2.37:
1
147.48%, 128%, 123%
(closer)
Better acceleration,
higher speeds
Honda CR-V
2.786:1,1.614:1, 1.082:1,
0.773:1, 0.566:1 Final
Drive: 4.500
172.6%, 149%, 140%
(medium)
Better acceleration,
higher speeds
Suzuki Swift Petrol
3.545:1, 1.904:1,
1.280:1, 0.914:1, 0.757:1
186%,148.75%, 140%
(wider)
Better acceleration from
stopped, better during
traffic
Engine Map
Automatic Transmission
Transmission
Engine
Automatic Transmission
• At zero speed ration the torque ratio will be several times higher
• As speed builds transmission approaches engine speed and torque drops
Automatic Transmission
Automatic Transmission Characteristics Curves.
Continuously Variable Transmission (CVT)
Chain Drive CVT
CVT
(a) flat disc and roller variable-speed drives
(b) toroidal disc and rollers variable-speed drives
Infinitely Variable Transmission (IVT)
Effect of Drive Train on Acceleration
• Estimation of acceleration performance requires modeling the
mechanical systems by which engine power is transmitted to
the ground.
Effect of Inertia
Te
T
E
IE
IC
D.S
IT
Axle
F.D
TDS
IFD
Inertia acting in opposite direction of rotation
Net Torque = Te- (Sum of inertia )
IAxle
IW
Torque @ Clutch
Torque output to Drive Shaft
Ta = Torque on the axles
Torque delivered to axles to accelerate rotating wheels and provide tractive force
• amplified by final drive ratio
• reduction due to inertia
Ta = Torque on the axles
Fx = Tractive force at the ground
r = Radius of the wheels
Iw = Rotational inertial of the wheels and axles shafts
w = Rotational acceleration of the wheels
Id = Rotational inertia of the drive shaft
d = Rotational acceleration of the drive shaft
Nf = Numerical ratio of the final drive
Effect of Drive Train on Acceleration
Fx 
Te N tf tf
r
 {( I e  I t ) N tf2  I d N 2f  I w }
ax
r2
g
a x  [ Fx  Rx  DA  W sin   Rhx ]
W
Te N tf tf
a
Ma x  [
 {( I e  I t ) N tf2  I d N 2f  I w } 2x  Rx  DA  W sin   Rhx ]
r
r
Fx= Max
Te N tf tf
ax
2
2
Ma x  {( I e  I t ) N tf  I d N f  I w } 2  [
 Rx  DA  W sin   Rhx ]
r
r
Te N tf tf
ax  [
 Rx  DA  W sin   Rhx ] /( M  M r )
r
Ta = Torque on the axles
Fx = Tractive force at the ground
r = Radius of the wheels
Iw = Rotational inertial of the wheels and axles shafts
w = Rotational acceleration of the wheels
Id = Rotational inertia of the drive shaft
d = Rotational acceleration of the drive shaft
Nf = Numerical ratio of the final drive
Effect of Power To Weight Ratio on Acceleration
Performance
• The ratio of engine power to vehicle weight is the first-order
determinant of acceleration performance.
Fx
g P
ax 
 1000
m / s2
M
VW
P  Engine Power in kW
W  Weight of the Vehicle in N
Velocity and Weight in denominator
• Acceleration capability decreases with increasing vehicle speed.
• Heavy trucks will have much lower performance levels than cars
because of the less favorable power-to-weight ratio
Velocity graph for different Power-to-Weight Ratio
0.16
0.18
0.19
0.21
Power-to-Weight Ratio Acceleration Time
Power Train Layouts-Passenger Cars
Front Engined Rear Wheel Drive (RWD)
Power Train Layouts-Passenger Cars
Front Engined Front Wheel Drive (FWD)
Power Train Layouts-Passenger Cars
Rear Engined Rear Wheel Drive
Four Wheel Drive
General arrangement of a part-time four-wheel drive system for off-road vehicles
Full Time Four Wheel Drive
Power Train Layout –Commercial Vehicles
Provides low flat floor area, great
choices for placing the doors, Good for
double deckers
Tandem Axle Drives for Heavy Vehicles
Reasons for Different Layouts
• Interior Space
– Front Engined, Transversely Mounted Front
Wheel Driven Vehicle will have more
passenger and luggage space, excellent for
small and medium size cars
• Ease of Handling
– Front Engine-RWD is good from the point of
view of handling, directional stability and
parking effort
– Front Engine-FWD handling behaviour is
comparatively low, higher parking effort,
Driver need to have good steering reflex
– Handling and Directional stability are poor
– Oversteer and understeer effects
Reasons for Different Layouts
• Maintaining Traction
– Traction Force is a product of Load acting on driven wheels and coefficient of friction, hence front engined front wheel driven vehicle and
rear engine rear wheel driven vehicle have better traction
– Acceleration performance of Front Engined RWD and Rear engine
RWD will be good due to dynamic load transfer
Reasons for Different Layouts
•
Balanced Braking
– Load Transfers, brake force and braking force play an important role during
braking
Front wheel driven
Rear Engine RWD is preferred for balanced braking
Rear Wheel Driven
Reasons for Different Layouts
•
Other Factors
– Engine Cooling
– Space Heating
– Noise
Comparisons of Layouts
Front and Rear Drive Acceleration Performance
Tractive Force
Tractive Effort 
Te N tf tf
r
W= Wstatic + Wdynamic longitudinal +Wdynamic transverse
W

FX
Tractive forces acting on a wheel.
Traction Limited Acceleration
• Assume there is adequate power from the engine, the
acceleration may be limited by the coefficient of friction
between the tire and road.
• In that case Fx is limited by:
Fx  W
  Peak coefficien t of Friction
W  Weight on drive wheels
• The weight on a drive wheel then depends on the static
plus the dynamic load due to acceleration, and on any
transverse shift of load due to drive torque
Transverse Weight Shift due to Drive Torque
• Transverse weight transfer occurs because the driveshaft
torque (Td) reacts between the frame mounted engine and the
axle. The engine, which is a part of the sprung weight,
produces the torque through the transmission. The rear axle
has to resist that torque at the rear tire patch.
– Transverse weight transfer adds cross weight under
acceleration and removes cross weight during deceleration.
– The magnitude of the transverse weight shift is a function
of the instantaneous engine torque and the roll stiffness
(K).
Transverse Weight Shift due to Drive Torque
• The driveshaft imposes a torque Td on the axle
• As the chassis roll compressing and extending springs on
opposite sides of the vehicle such that a torque due to
suspension roll stiffness Ts is produced
At equilibrium
Now Td can be determined by the following equation
To find Ts roll torque produced by suspension requires whole vehicle analysis
•Drive torque reaction at engine /transmission is transferred
and distributed between front and rear suspension
• Roll torque produced by suspension is proportional to roll angle
Drive Torque Reactions
• Roll torque produced by suspension is
proportional to roll angle
Roll angle= drive torque/roll stiffness
Substituting Ts and Td values in the above equation
During Acceleration
Substitute for ax
Solving for Fx
Traction Limits
• The maximum tractive force that can be developed by a solid
rear axle with a non-locking differential:
Wb
L

h
2 r Kf
1  
L
N f t K

Fx max
• For a solid rear axle with a locking differential, additional
tractive force can be obtained from the other wheel up to its
traction limits such that the last term in the denominator of the
above equation drops out.
Wb

L
Fx max 
h
1 
L
• The above equation holds good for independent rear suspension
Traction Limited Acceleration
• The maximum tractive force that can be developed by a solid front
drive axle with a non-locking differential:
Fx max
Wc

L

h
2r Kf
1  
L
N f t K
• For a solid front drive axle with a locking differential, or
independent front suspension
Wc

L
Fx max 
h
1 
L
• The load on the front axle will be higher than rear axle,
however,the load gets transferred during acceleration.
• The acceleration performance of front wheel drive vehicle will not
be proportionately better compared to rear wheel drive vehicles,
though numerical values of performance is higher
Locking Differential
• A locking differential is designed to overcome the chief
limitation of a standard open differential by essentially
"locking" both wheels on an axle together as if on a common
shaft. This forces both wheels to turn in unison, regardless of
the traction (or lack thereof) available to either wheel
individually.
• A locked differential forces both left and right wheels on the
same axle to rotate at the same speed under nearly all
circumstances, without regard to tractional differences seen at
either wheel. Therefore, each wheel can apply as much
rotational force as the traction under it will allow, and the
torques on each side-shaft will be unequal.(Unequal torque,
equal rotational speeds)
Locking Differential
Acceleration
• Acceleration with mu=0.8
Acceleration with Mu=0.2
Acceleration with mu=0.05
Acceleration Performance
=1
 = 0.8
 = 0.6
 = 0.4
 = 0.2
Slip
Wheel
wr  V
Slip 
wr
100 % slip means wheels are spinning but vehicle is not moving
Road  will be optimum at a slip value of approximately 20%
Traction Control System
Traction Control
Demo
Traction Control
General Equation for Braking
Arbitrary forces acting on a vehicle.
• The general equation for braking performance may be obtained
from Newton’s Second Law written for the x-direction.
Fxt
Dx  ( Fxf  Fxr  DA  W sin  ) / M 
M
Constant Deceleration
• Assuming that the forces acting on the vehicle will be constant
throughout a brake application.
Fxt
dV

M
dt
Fxt  The Total of all longitudin al accelearti on forces
Dx 
V  Forward velocity
This equation can be int egrated for decelerati on
Vf
t
Fxt s
V dV   M 0 dt
o
V0  V f 
Fxt
ts
M
V0= Initial Velocity
Vf = Final Velocity
ts =Time for the Velocity Change
Time to Stop
The following equation gives the time required to bring the
vehicle from initial velocity to final velocity
Fxt
dV

M
dt
Fxt  The Total of all longitudin al accelearti on forces
Dx 
V  Forward velocity
This equation can be int egrated for decelerati on
Vf
t
Fxt s
V dV   M 0 dt
o
V0  V f 
Fxt
ts
M
Fxt
 (V0 V f ) / Dx
M
V f  0, then
t s  (V0  V f ) /
if
ts 
V0
Dx
Stopping Distance
Fxt
dV

M
dt
Fxt  The Total of all longitudin al accelearti on forces
Dx 
V  Forward velocity
In the case where the deceleration is
a full stop, then Vf is zero, and X is
the stopping distance, SD. Then:
This equation can be int egrated for decelerati on
Vf
V02
V02
SD 

Fxt
2 Dx
2
M
t
Fxt s
V dV   M 0 dt
o
dx
dt
dx
dt 
V
V
Vf
X
Fxt dx
dV



M 0 V
V0
V02  V f2
Fxt
x
2
M
V02  V f2 Fxt
x(
)/
2
M

Energy Absorbed by Brake System
•
•
The energy and/or power absorbed by a brake system can
be substantial during a typical maximum-effort stop.
The energy absorbed is the kinetic energy of motion for the
vehicle, and is thus dependent on the mass.

M 2
2
Energy 
V0  V f
2

Power Absorbed by Brake System
•
•
The power absorption will vary with the speed, being
equivalent to the braking force times the speed at any
instant of time.
Thus, the power dissipation is greatest at the beginning
of the stop when the speed is highest. Over the entire
stop, the average power absorption will be the energy
divided by the time to stop. Thus
M V02
Power 
2 ts
Other Braking Forces
• Rolling Resistance
• Aerodynamic Drag
• Driveline Drag
• Engine Drag- not much an effect
• Grade
Brakes
Disc Brake
Drum Brake
Drum Brakes Vs Disc Brakes
• Drum brakes have high brake factor and the easy
incorporation of parking brake features.
• On the negative side, drum brakes may not be as consistent in
torque performance as disc brakes.
• Disc brakes have lower brake factors, require higher
actuation effort, and development of integral parking brake
features has been required before disc brakes could be used at
all wheel positions.
• Disc brakes have better torque performance
Brake Factor
Brake Factor is a mechanical advantage that can be utilized in drum
brakes to minimize the actuation effort required
Brake Fading
Torque Vs Time.
The reduction of friction termed brake fade is caused when the temperature
reaches the "kneepoint" on the temperature-friction curve
Braking Fundamentals
W
WheelTorque
Brake Force =bPA
Brake Torque = bPAr
Direction of Travel
Brake
Torque
Braking Force =W
Wheel Locking
•
•
•
•
•
If brake force is greater than braking force wheel locks –panic
braking
Vehicle becomes unstable
A vehicle with locked front wheels will have a stable straightahead motion, however, the steerability is lost
A vehicle with locked rear wheels will be unstable; it turns
around and ends up in sliding
Usually, it is preferred that the front wheels lock first.
Wheel Lockup
Wheel Lockup
bpA <=W to avoid wheel lock up
To avoid wheel lock up
1. Control brake pressure p ----Proportioning
2. Optimal use of  between the road and tyre - ABS
3. Measure W on each wheel and also estimate , based
on that control oil pressure p -EBFD
Brake Proportioning
• Due to dynamic load transferring load acting on the axles
change and some times load acting on each of the wheels is
different
• Brake force applied to each one of the wheels must be
proportionate to the vertical load acting on wheel
• Distribution of the brake force at each of the wheels based on
the load acting on the wheels is called proportioning
• If proper proportioning is not done, there is a chance of
wheels getting locked up due to brake-forces
• The variation of forces on the front and rear axle during
deceleration and for different friction coefficient is given
below
Load Variation on Axles
• Variation of load on the axle during braking also with respect to
road friction
Braking System
Tractor/Trailer Air Brake System
Brake Circuit with Proportioning Valves
LSPV-Load Sensing Proportioning Valve
G-Valve: Load Sensing Valve (Deceleration)
87
Proportioning Valve
• The proportioning valve is designed to prevent rear wheel
lockup during panic braking.
• It is needed for two main reasons - most vehicles use drum
brakes in the rear which are hydraulically operated and all
vehicles experience weight transfer during a panic stop
• The most important point to understand about proportioning
valves is when they work - only during panic braking. This
means that a vehicle could potentially go its whole life
without ever using its proportioning valve
• Proportioning Valves
– Pressure Sensitive Proportioning Valve
– Load Sensing Proportioning Valve
Concept of Load Sensing Proportioning Valve
Rear of
Vehicle
Frame Position when Braked
Oil to Rear
Brake Cylinders
With Pressure p
Initial Frame Position
Front of
Vehicle
Oil From
Master Cylinder
With Pressure P
1. Deceleration Vs Vehicle Lift -characteristic required
2. Vehicle Lift Vs Oil Pressure characteristic required so that valve
movement can be fixed
Proportioning Valve
Proportioning Valve
Federal Requirements for Braking Performance
• Out of the public concern for automotive safety in the 1960s,
the Highway Safety Act of 1965 was passed establishing the
National Highway Traffic Safety Administration charged with
promulgating performance standards for new vehicles which
would increase safety on the highways.
• Among the many standards that have been imposed are Federal
Motor Vehicle Safety Standard (FMVSS)
• FMVSS 105, establishing braking performance requirements
for vehicles with hydraulic brake systems,
• FMVSS 121, establishing braking performance requirements
for vehicles with air brake systems.
Federal Requirements for Braking Performance
1)
2)
3)
4)
5)
Although the standard is quite detailed and complex, the requirements for
stopping distance performance can be summarized into tests:
First effectiveness – A fully loaded passenger car with new, unburnished
brakes must be able to stop from speeds of 30 and 60 mph in distances that
correspond to average deceleration of 17 and 18 ft/sec2, respectively.
Second effectiveness – A fully loaded passenger car with burnished brakes
must be able to stop from 30, 60 and 80 mph in distances that correspond to
average decelerations of 17, 19 and 18 ft/sec2.
Third effectiveness – A lightly loaded passenger car with burnished brakes
must be able to stop from 60 mph in a distance that corresponds to an average
deceleration of 20 ft/sec2.
Fourth effectiveness – A fully loaded passenger car with burnished brakes
must be able to stop from 30, 60, 80 and 100 mph in distances that correspond
to average decelerations of 17, 18, 17 and 16 ft/sec2, respectively.
Partial Failure – A lightly loaded and fully loaded passenger car with a failure
in the brake system must of able to stop from 60 mph in a distance that
corresponds to an average deceleration of 8.5 ft/sec2.
Load on Front and Rear Axles During Deceleration
c
hW
W
Dx  W fs  Wd
L
L g
b
hW
Wr  W 
Dx  Wrs  Wd
L
L g
W fs  Front axle static load
Wf 
Wrs  Re ar axle static load
 h  W
Wd   
 L  g

 Dx  Dynamic load transfer

hW
Dx )
L g
hW
Fxmr   pWr   p (Wrs 
Dx )
L g
 p  Peak coefficien t of friction
Fxmf   pW f   p (W fs 
Proportioning Line
Dx 
( Fxmf  Fxr )
M
Fxmf 
for Fxmr
Dx 
( Fxmr  Fxf )
M
Fxmr
 pW fs
h
1  p
L
 pWrs

h
1  p
L
p
h

Fxr
h L
1  p
L
p h

Fxf
h L
1  p
L
Proportioning line for Different Mu and FMVSS Conditions
Antilock Braking System
Slip
• Wheel locking condition is expressed by 100% slip
• Slip is defined as =
V  r
V
– V: vehicle Velocity, : Tire rotational speed
Braking Coefficient
• Braking Coefficient = Braking Force/Tire load
• Braking force will be large if braking coefficient is high which
can happen at 20% wheel slip for all types of road
• Braking coefficient is very low at 100% slip, wheels are not
spinning, means vehicle skids
Improving Braking Force
• To have better braking the wheel speed should be cyclically
altered between point 2 and 3
Wheel Speed Cycling
•
Figure shows wheel speed with respect to time when brake
is applied
•
When the brakes are first applied, wheel speeds diminish
more or less in accordance with the vehicle speed in region
1 in the plot
Algorithm for Pressure Modulation
1.
Brake pedal
2.
Brake servo unit
3.
Master cylinder with
brake-fluid reservoir
4.
Brake
5.
Wheel-speed sensor
6.
Warning lamp
Complete ABS System
Braking System with ABS
ABS Variants
Control channel
Wheel-speed sensor
Wheel-speed sensor
(alternative too
differential speed
sensor)
Front wheels
Rear wheels
ABS Variants
Control channel
Wheel-speed sensor
Wheel-speed sensor
(alternative too
differential speed
sensor)
Front wheels
Rear wheels
Working of ABS System
Truck Hydraulic Braking System and ABS
ABS for Trucks
ABS for Buses
Electronic Brake Force Distribution-EBD
• Proportioning can be done by using mechanical brake
proportioning valves
• Modern practice is to use Electronic Brake Force
Distribution (EBFD) technology
• EBFD is an automobile brake technology that
automatically varies the amount of force applied to each
of a vehicle's brakes, based on road conditions, speed,
loading, etc.
• EBD can apply more or less braking pressure to each
wheel in order to maximize stopping power
Electronic Brake Force Distribution
Compute by sensing V and 
V  r
slip 
V
Compute load on the wheel by sensing
deceleration
Compute =W on each wheel, control oil
pressure to control brake force
2009 Corolla ABS, EBD and Brake Assist Video
Braking Efficiency
• Braking efficiency = a property defined for a complete vehicle
• Braking efficiency = deceleration/(g*max(braking coefficient))
• It can be seen as totally used friction (deceleration/g) in relation
to how well the best used axle (or wheel) uses the friction if
100%, the distribution of braking is optimal.
b 
•
Dact
p
The braking efficiency concept is useful as a design tool for
the designer to assess success in optimizing the vehicle
braking system.
Braking Efficiency
Fifth Gear ABS, TCS and ESP video