It is obvious that vehicle weight has a linear relationship with the energy to be dissipated (stored) and the change in velocity required has a exponential relationship. • Deceleration times and stopping distances vary somewhat for all vehicles on a given road surface. • It should then be obvious that sizing the brake system components has critical importance with respect to the potential vehicle velocity and the mass of the vehicle. • Note that heavy trucks generally have greater stopping distances as compared to typical passenger cars.

1. Braking Performance
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2. Stopping Distance
• It is the distance between
the moment when a hazard
is recognized and the time
when the vehicle comes to a
complete stop.
• It is the sum of the distance
traveled during the reaction
time at given speed and
active braking time.
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3. Reaction Time
• The reaction time is the period which elapses between
the recognition of the object, the decision to brake, and
the time it takes for the foot to hit the brake pedal.
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4. Reaction Time
• The reaction
time is not a
fixed value: it
ranges from
0.3 to 1.7 s,
dependingdepending
upon the driver
and on
external
factors.
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5. Stopping Sight Distance (SSD)
• Worst-case conditions
– Poor driver skills
– Low braking efficiency
– Wet pavement
• Perception-reaction time = 2.5 seconds
• Equation:
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rtV
a
V
SSD 1
2
1
2
+=

6. Speed and Stopping Distance
• Look carefully at stopping distance for each vehicle
speed !!!
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7. Vehicle Braking
• It is obvious that vehicle weight has a linear relationship
with the energy to be dissipated (stored) and the change
in velocity required has a exponential relationship.
• Deceleration times and stopping distances vary
somewhat for all vehicles on a given road surface.somewhat for all vehicles on a given road surface.
• It should then be obvious that sizing the brake system
components has critical importance with respect to the
potential vehicle velocity and the mass of the vehicle.
• Note that heavy trucks generally have greater stopping
distances as compared to typical passenger cars.
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8. Braking and Deceleration
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9. Retarding Forces
• The retarding forces on a vehicle are generally of the
type shown in the following equation, where q is uphill
grade:
– Front and rear brake forces
– Aerodynamics drag– Aerodynamics drag
– Rolling resistance
– Climbing resistance and downgrade force
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sinθWFFFFD
g
W
aM RollingcAerodynamirBrake,fBrake,xx -----=÷÷
ø
ö
çç
è
æ
-=
FBf , FBr include: 1) brake torque
2) bearing friction
3) drive line drag.

10. Rolling Resistance
• It is the product of deformation processes which occur at
the contact patch between tire and road surface.
• Frr = f Mg
Depends on:
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10 ME 467 Vehicle Dynamics
Depends on:
– bearing friction,
– deformation of tires,
– road surface.
Rxf + Rxr = fr (Wf + Wr) = frW
fr : rolling resistance coefficient.
Rolling resistance: depends on the
distribution of loads on axles.
: it equivalent to about 0.01 g
(0.3 ft /sec2).

11. Rolling Resistance
• It is directly proportional to
the level of deformation and
inversely proportional to the
tire radius.
• It increase with greater
loads, higher speeds, and
lower pressure.
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12. Aerodynamic Drag
• It is empirical value and
depends on vehicle
shape and speed.
Fa = CdV2
Depends upon:
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Depends upon:
– frontal area,
– relative vehicle speed to air speed
DA = C . A . V2
DA: can be neglect at low speed,
DA » 0.03 g at high speed (1 ft /sec2).
A: frontal area
V: relative speed
C: aerodynamic drag coefficient.

13. Drive line drag (inertia)
• Inertia (adds to the effective mass of vehicle).
• Drag arises from internal friction in gears
and bearings and from engine braking.
• Engine braking arising from internal friction
and air pumping losses.and air pumping losses.
• Engine braking multiplies by the gear ratio
selected.
• If vehicle decelerates faster than drive line,
slowing down from drive line drag will have
lower contribution in braking effort.
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14. Grade
Rg = W sinθ
Grade increase the braking effort of vehicle in uphill
road and decrease it in down hill road.road and decrease it in down hill road.
Rg = W sinθ = W θ , θ in (radians) ≈ grade for small angles.
Grade of 4% (0.04) will be equivalent of deceleration
(! 0.04g) →1.3 ft /sec2
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15. Active Braking and Kinetic
Energy
• Brake systems convert the kinetic energy of the vehicle
at velocity to some other form of energy.
• Current design brake systems in conventional
automobiles convert the kinetic energy of the vehicle into
thermal energy (heat) which is subsequently dissipated
into the atmosphere.into the atmosphere.
• Hybrid vehicles attempt to recover the kinetic energy as
stored battery or capacitor electrical energy.
• Kinetic energy (KE) is mathematically represented as ½
the product of the vehicle mass and the velocity of the
vehicle squared.
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15
ME 467 Vehicle Dynamics
2
2
1
vmKE =

16. Energy Conversion during
Braking
Conversions that might be possible would include:
Kinetic Energy to Thermal Energy
• Typically a friction system generating heat between a rotating
wheel part and a vehicle fixed friction component which can
be modulated by the driver.be modulated by the driver.
Kinetic Energy to Stored Electrical Energy
• Regenerative system where electric motors can be used as
electric generators under braking and subsequently turn some
portion of the kinetic energy into stored electrical energy.
• The stored electrical energy can be later used to supplement
performance and/or range.
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17. Energy Conversion during
Braking
Kinetic Energy to Stored Mechanical Energy
• Regenerative system where mechanical systems can be engaged
under braking to impart kinetic energy to a flywheel subsequently
turning some portion of the kinetic energy into stored mechanical
energy.
• The stored energy can be released as required to augment
acceleration and/or extend the range of the vehicle.acceleration and/or extend the range of the vehicle.
Kinetic Energy to Stored Hydraulic Pressure Energy
• Regenerative system where hydraulic motors can be used as
hydraulic pumps under braking and subsequently turn some portion
of the kinetic energy into stored hydraulic energy.
• Storage of the fluid under pressure can later be released to the
pump, driving it as a hydraulic motor which can be used to augment
acceleration and/or extend the range of the vehicle.
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18. Constant Deceleration
• Assuming the forces on the vehicle are constant:
• This equation can be integrated because F is constant
dt
dV
M
F
=D
xtotal
x =
• This equation can be integrated because Fx is constant
from initial velocity Vo to final velocity Vf:
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òò =
sf
0
t
0
V
V
xtotal
dt
M
F
dV

19. Constant Deceleration
• The fundamental relationship equations that govern
braking are:
òò =-=>=
t
s
xtotal
f0
V
xtotal
t
F
VVdt
F
dV
f
Where ts is the time for the velocity change
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òò =-=>=
0
sf0
V
t
M
VVdt
M
dV
o

20. Stopping Distance (SD)
• can substitute for dt in equation (1) and integrate to
obtain the relationship between velocity and distance:
x
M
F
2
VV xtotal,
2
f
2
0
=
-
Where x = distance traveled during deceleration
• In the case where Vf = 0, then x = stopping distance
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M2
x
2
o
D2
V
x =

21. Braking Energy
• The time to stop is then:
• Thus, all things being equal, the time to stop is
x
0
xtotal
0
s
D
V
M
F
V
=t =
• Thus, all things being equal, the time to stop is
proportional to the velocity, where as the distance to stop
is proportional to the velocity squared.
• The energy absorbed by the brake system is the kinetic
energy of motion:
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)V(V
2
M
=Energy 2
f
2
0 -

22. Deceleration with Wind
Resistance
• Aerodynamic drag effects:
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23. Types of braking systems
• Desk and Drum types
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24. Physics of braking
•d
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25. 3/27/2012 25 ME 467 Vehicle Dynamics

26. Brakes
Drum brakes
Advantage
- High brake factor (low actuation effort).
- Easy to integrate with park brake.
Disadvantage
- not consistent in torque performance.- not consistent in torque performance.
Disc brakes
Advantage
- more consistent torque
Disadvantage
- low brake factor.
- high actuation effort.
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27. Brake factor
Is a mechanical advantage in drum brakes to minimize required
actuation effort?
Moment equation about pivot of shoe A
åMp = e Pa + n m NA – m NA = 0
Pa : application of an actuation force.
NA : normal force
e, n, m: distances as shown in (fig).
A
e, n, m: distances as shown in (fig).
NA = Pa · e , NB = Pa · e
m - m n m + m n
Friction forces on brake shoes:
FA = m NA , FB = m NB ,
FA = Pa · m · e , FB = Pa · m · e
m - m n m + m n
FA = m e , FB = m e
Pa m - m n Pa m + m n ٢٧

28. FA = m e , FB = m e
Pa m - m n Pa m + m n
TL = FA · r TT = FB · r
TL : brake torque of leading shoe,
TT : brake torque of trailing shoe.
Braking torque developed by leading shoe greater than brake
torque developed by trailing shoe.
The moment produced by the friction force on the leading shoe acts to rotateThe moment produced by the friction force on the leading shoe acts to rotate
against the drum and increase the friction force developed (self servo).
For this reason two leading shoe type of brake usually using
in front brakes.
In leading shoe If µ gets too large ≈ 1 brake factor goes to
infinity ( m ≈ n) and brake will lock.
Consequences:
1- sensitivity to the lining coefficient of friction
2- high noise ٢٨ME 467 Vehicle Dynamics

29. Leading show A :- Moment produced by the friction force on the
show acts to rotate it against the drum and increase the friction force
developed (self-servo which characterized by brake factor).
Trailing show B :- friction force acts to reduce the application
force, lower brake factor and higher braking forces required to achieve
desired braking torque.
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30. Braking torque during stopping processBraking torque during stopping process
Tb = f ( Pa , velocity , temperature )
•Torque normally increase linearly with Pa
•Torque increases as velocity increase.
•Torque decreases as temperature increase.
Disk brake shows less torque variation during stopping.
So, in drum brakes it is difficult to maintain proper balance
between front and rear braking effort during max braking.
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31. Braking force at the ground
Fb = Tb – Iw aw
r
That if wheels still rotating during braking
Iw : rotational inertia of wheels
aw: rotational deceleration of wheels.aw: rotational deceleration of wheels.
If wheels lock up:
aw related to acceleration of vehicle
aw = ax ,
r
Iw: part of vehicle mass
Fb = Tb
r
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32. road friction)-Traction (Tire
Friction force can be increase to
the limit of frictional coupling
between the tire and road
Mechanisms of friction:
• Adhesion, it arises from
intermolecular bonds between
the rubber and road surface.
Higher on dry road, reduced
with wet road surface
(wheel rotating).
• Hysterises, Represents energy
loss in the rubber as it deforms
when sliding over road surface
not affected by water, thus it
has better wet traction. ٣٢

33. - Both Mechanisms of
friction depend on small
amount of slip occurring in
the contact patch.
- As braking force develops,
additional slip is observedadditional slip is observed
as a result of deformation of
the rubber elements.
- Deformation increases
from front to back and the
force developed increases
proportionality.
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34. Both mechanisms depends
on two factors:-
(brake force coefficient and
slip - which are coexistent)
1) slip of tires
slip = v – w . r , %
v
v : forward velocityv : forward velocity
w: tire rotational speed
w¯….slip- ,
w=0….slip=100%
2) brake force coefficient (µ)= Fx
Fz
Fx: braking force,
Fz: axle load
Brake coefficient µ increases with slip to
about 10-20% slip (see fig.)
µp – peak coefficient (it establishes the
max brake force that can be obtained
from particular tire and road friction).
Max braking coefficient develops at µp .
Then if Slip- … µ ¯ until µs .
ms … at full wheel locking.
in ABS ms return to mp in a repeated
cycles.
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35. Friction in braking depends on:
1. tire design
2. road surface type
3. velocity (both peak and slide friction decrease with
velocity increase).
4. Inflation pressure. Wet road: -P ® -mp ,ms
Dry road: P less affect on mDry road: P less affect on m
5.vertical load: (Brake force coefficient=Fx /Fz ),
increasing of vertical load reduce brake force coefficient levels,
if load increase mP ,ms don’t increase proportionally.
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36. Brake proportioning: describes the
relation between front and rear brake
forces, determined by the pressure
applied to each brake wheel cylinder
and the gain of each.and the gain of each.
Ideal design is to bring both axles up to
lock up simultaneously
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37. Brake proportioning
Dynamic load during braking
W , W : static loadsWfs , Wrs : static loads
Wd : dynamic load transfer.
Maximum brake forces on an axle:
mp- peak coefficient of friction ٣٧

38. Fxm = f(m , Dx)
(see the fig)
Substituting Dx in previous
eq. of Fxmf, Fxmr (shown
again below) yields the
equations in the square
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39. The intersection point can
be determined by
manipulating equations of
Fxmf, Fxmr.
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40. Brake gain and Brake proportioning
The gain of brakes on front and
rear wheels is an important
factor to determine brake
proportioning.
Brake force on each wheel
Fb =Tb =G Pa
r rr r
G: brake gain (in–lb/psi)
Pa: application pressure.
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41. Hydraulic proportioning valve
Hydraulic proportioning valve: provide equal pressure
to both front and rear brakes up to certain pressure
level and then reduce pressure to rear brake.
So, it adjusts the brake torque output on front and rear wheels in
accordance to the peak traction forces possible or to loads.
Example:- Hydraulic proportioning valve identification
number:(500 / 0.3) that means:-
For Pa < 500 psi
Pf = Pr = Pa = application pressure.
For Pa > 500 psi
Pf = Pa , Pr = 500 + 0.3 (Pa – 500)
If Pa = 700 psi
Pf = Pa = 700 psi
Pr = 500 +0.3(700 – 500) = 570 psi.
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