Brakes

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Brakes

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Legislative service braking

0,540,36(ta+ ts/2) [s]

700500F [N]

5,05,8a [m/s2]

36,761,236,750,7s [m]

s [m]

60806080v [km/h]

N3N2N1M3M2M1Category

1501,0

2v

v+130

15,02

vv+

Requirements on the brake capability of vehicles stated in regulations:

ECE 13, ES 71/320 (For CZ: . 102/1995 Sb.)

Braking test with cold brakes, whennoengine brake is applied for

empty and fully loaded vehicle

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Legislative emergency braking

2,2

600400Fhand[N]

700500Ffoot[N]

2,52,9a [m/s2]

71,6123,364,493,4s [m]

s [m]

60806080v [km/h]



ECE 13, ES 71/320 (For CZ: . 102/1995 Sb.)

Braking test with cold brakes, whennoengine brake is applied for

empty and fully loaded vehicle

15021,0

2v

v +130

215,02

vv +

115215,0

2vv +

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Legislative partial failure

performance

A maximum pedal force of approx. 445 N shouldachieve deceleration of approx. 0,3 g for the vehicleloaded at GVW in the event of booster failure.

In case of hydraulic circuit failure, a maximum pedalforce of approx. 445 N should slow the vehicle loadenat GVW at decelaration of approx 0,3 g.

In the event of repeated or continued braking withincreased brake temperatures a pedal travel ofapprox. 115 to 130 mm out of 150 mm availableshould be exceeded for a maximum pedal force ofapprox. 445 N.


Federal Motor Vehicle Safety Standard (FMVSS) 105

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Legislative Parking brake

The parking brake should hold the vehicle stationarywhen laden at GVW on a 30 % slope with a hand

force of not more than 356 N or a foot force of less

than 445 N.

With the apply force limitations stated, the parking

brake should be able to slow a vehicle laden at GVW

at approx. 0,3 g.


Federal Motor Vehicle Safety Standard (FMVSS) 105

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Braking procedure

tr driver reaction time

ta brake system application time

ts deceleration rise timetv constant deceleration time

s1 distance traveled during reaction and

system application time

s2 distance traveled during deceleration risetime

s3 distance traveled during constant

deceleration interval

Source: Limpert R.

Brake design and Safety

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Driver reaction time

Perception phase: 0,32 to 0,55 s

Judgement and reaction initiation: 0,22 to 0,58 s

Reaction execution (pedal switch): 0,15 to 0,21 s

Driver reaction consists of:

perceptionjudgementreaction initiationreaction execution

Drivers use distributiveattention to scan entire scene. After swith to

concentrativeattention the controlled reaction begin.

Source: Limpert R.


1,480,771,25Eye movement > 5o

1,330,681,12Eye movement 0,5 to 5o

0,780,360,64No eye movement

98 %

(only 2 % are slower)

2 %

(only 2 % are faster)Normal driverReaction time [s]

Reaction time

Source: Mitschke M., Wallentowitz H.

Dynamik der Kraftfahrzeuge

The reaction time under influence of alcohol is multiplication of the standard driver reaction time

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Brake system application time

Deceleration rise time

0,180,140,17Deceleration rise time

passenger cars (ts)

0,060,030,05Brake system application

time passenger car (ta)

98 %

(only 2 % are slower)

2 %

(only 2 % are faster)

Normal

passenger carVehicle

Reaction time [s]

0,540,36

(ta+ ts/2) [s]






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Braking procedure

( )

2max

max

21

1321

max1

2

2

max2

1

maxmax

2

2

2max2

0

3

max

1

max

2

max12

max2max2

2max1

0

2

2max1

max1

max

1

2422

42

1

2

2

2

2

6

2

ss

ar

ss

vv

t

sv

s

ss

t

ss

s

artravel

ta

a

vtttvssss

tavta

vaa

v

tatvdtvs

t

a

v

a

vt

ta

vv

tavdtavv

ta

tvdtvs

t

t

avdtt

t

avv

tt

ax

ttvs

v

s

+

++=++=

++

=

=

=+==

=

=

+=

+=+=

+==

+=+=

=

+=

&&

Source: Limpert R.


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Change of kinetic energy into the

heat

=

+=

+

++=

+++=

dtPW

PP

P

vScg

xsfvGvvSc

g

xGsGfGP

R

drivetrain

brakeengine

xxxxR

_

32

22

&&&&

Necessary power on the wheels [W]

Necessary braking power, the resulting power

after engine brake is applied and when vehicle

resistances are acting on the vehicle [W]

Braking work [J] dissipated energy which is changed into the heat in the

vehicle brakes

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Design of vehicle brake system

http://static.howstuffworks.com/flash/brake-simple.swf

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Disc brakes

The rotor (disc) rotates through the

caliper. The wheel cylinder pistons

force the braking pads against the

disc.

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Disc brakes

Fixed caliperCaliper solidly bolted to the flange.

Two or four pistons. Pistons from both sides of the disc

Source: Limpert R.


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Disc brake

Floating caliper

http://static.howstuffworks.com/flash/disc-brake.swf

One or two pistons on inboard side only.

The pressure forces the piston and pad toward the disc and also forces the

housing in the opposit direction (to apply the outboard pad against the disc).

Source: Limpert R.


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Fixedversusfloating caliper

Fixed caliper

more balanced inner and outer

pad wear

no anchor or general knuckle

attached with standard fasteners

fewer service parts

Floating caliper

easier to package

lower brake fluid operating

temperature

fewer leak points

easier to bleed in service

Air-disc floating caliper

Rockwell International

Source: Limpert R.Brake design and Safety

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Wear in disc brakes

Worn disc brakes can show significantly morewear on the leading end (rotor entrance) than on

the trailing end (rotor exit). The nonuniformdistributrion is caused by the lever arm betweenthe pad drag force and abutment force.

Solution to minimize or eliminite tapered padwear can be: Asymmetrical caliper piston contact edge

Piston located closer to the trailing edge

Four pistons per caliper (smaller piston located at theleading end)

ITT Teves Hammerhead design of pad anchorsystem

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Wear in disc brakesSource: Limpert R.Brake design and Safety

Non uniform pressure distribution wears the brake pads unevenly,particularly during severe braking in high speeds.

Pad wear increases for brake temperature in excess of approx. 573 to623 K.

Uniform pad wear is a major indicator of a quality of caliper design.

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Non-uniform pad presure distribution

Source: Limpert R.


Fav average force pressing the pad against

the rotor (N)

lp pad length (mm)

tp pad thickness (mm)

f pad support friction coefficientp pad/rotor friction coefficient

Equation of moment equilibrium around point A

62

pp

fpavppav

lFl

FtF

=+

+

=

2

6p

fppp

p

av lt

l

FF

++=

2

61max

p

fppp

p

av

lt

lFF

Solving for force change

Maximal forceSolving for typical values leads to resultthat the force (pressure at the leadingedge will be one third greater thanthe average force, and only two thirds

at the rotor exit.

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Offset piston design

Source: Limpert R.



Equation of force equilibrium

Both equations of equilibrium combined together tosolve c

2

p

fpdav

lFtFcF +=

fpppav FFF +=

fp

p

fppp

lt

c

+

+=

1

2

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Hammerhead pad designSource: Limpert R.


6

p

fppav

lFbFtF

=

( )btl

FF fpppp

av = 6

( )

+= bt

lFF fppp

p

av 6

1max


Solving for force change

Maximal force

Solving for typical values leads to result that the maximal force Fmax= 1,033*Fav. Which shows

nearly uniform distribution. Pulled pads can carry heavier specific loadings and are used in

high performance vehicles.

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Proposal of piston diameter

Higher pressuresmaller components of brake system

Higher pressurehigher demands on sealing

From experience: optimal pressure in braking system

for z = 1, p = 100 bar (1000 N/cm2)

Cirfumferential force acting on the disc:

BhydpistonU pAF =

*

_

*

*

2

CpArM

CpAF

C

hydpistondiscBB

hydpistonU

B

=

==

Two friction areas, and corresponding change of circumferential force:

Friction torque of disc brake

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Proposal of piston diameter

Braking forces on the vehicle wheels

( )

DB

G

p

z

B

G

DBBpz

GBBp

GBBz

B

BDB

CArr

B

CArr

B

CpArr

B

CpArr

B

hydF

Fhyd

RFhyd

RF

F

R

RRpistonRdiscB

dyn

R

FFpistonFdiscB

dyn

F

RhydRpistonRdiscB

dyn

R

FhydFpistonFdiscB

dyn

F

+=

+=

+=+=

=

=

=

=

=

1

1

2

2

2

2

*

*

**

*

*

*___

*

*

___

*

*

___

*

___

Characteristic parameter (front brakes)

Characteristic parameter (rear brakes)

Brake force distribution factor

Estimation of front brake characteristicparameter

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Source: Audi A4

ATZ Sonderheft, 2008

Electromechanical Park brake acting

on disc brake

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Master cylinder

The cars brake systems are split intotwo circuits, with

two wheels on each circuit. If a fluid leak occurs in one

circuit, only two of the wheels will lose their brakes and

your car will still be able to stop when you press the brake

pedal.

Themaster cylindersupplies pressure to

both circuits of the car. When the brake

pedal is pressed, it pushes on theprimary

pistonthrough a linkage. Pressurebuilds

in the cylinder and lines as the brake pedalis depressed further. The pressure between

the primary andsecondary pistonforces

the secondary piston to compress the fluid

in its circuit. If the brakes are operating

properly, the pressure will be the same in

both circuits.

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Master cylinder

Estimation of piston diameter

( )

z

F

G

DBBiA

FA

ip

ApF

FiF

BpF

Bpcylindermaster

Bp

cylindermaster

Bp

hyd

cylindermasterhydcircuit

BpBpcircuit

+

=

=

=

=

1*

1__

1__

1__1_

Brake system without booster

FBp Force on brake pedal

iBp Lever ratio of brake pedal

Brake system with booster

( )

( )

( ) ( )Gz

DBBFFi

A

ADBB

GzApF

FFiFFF

DBB

Gzp

FspringboosterBpBp

cylindermaster

cylindermaster

F

cylindermasterhydcircuit

springboosterBpBpspringboostercircuitcircuit

F

hyd

+

=

+

==

==

+

=

1

1

1

*

_

1__

1__*1__1_

__1_

*

Fbooster_spring Force of booster

return spring

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Booster

Thevacuum boosteris a metal canister that

contains a valve and a diaphragm. A rod going

through the center of the canister connects to the

master cylinder's piston on one side and to the

pedal linkage on the other.

The engine creates a partial vacuum

inside the vacuum booster on both

sides of the diaphragm. When the

brake pedal is hited, the rod cracks

open a valve, allowing air to enter thebooster on one side of the diaphragm

while sealing off the vacuum. This

increases pressure on that side of the

diaphragm so that it helps to push the

rod, which in turn pushes the piston inthe master cylinder.

As the brake pedal is released, the

valve seals off the outside air supply

while reopening the vacuum valve. This

restores vacuum to both sides of the

diaphragm, allowing everything to

return to its original position.

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Booster

Source: Limpert R.


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Estimation of the booster diaphragm

area

( )

( )

( )

( )

( )

+

+

=

+

==

+

+

+

=

+=

+=

=

+=

=

BpBpspringboosterF

cylindermaster

Stbooster

F

hyd

cylindermaster

springboosterboosterStBpbooster

springbooster

F

cylindermaster

BpBp

booster

F

cylindermaster

springboosterBpBpbooster

cylindermaster

springboosterboosterStBpBpbooster

hyd

cylindermaster

springboosterBpBpboosterhyd

springboosterboosterStBpboosterStcircuit

springboosterBpBpboostercircuit

FiFDBB

AGz

pA

DBB

Gzp

A

FApFi

FDBB

GAz

Fii

G

DBB

A

FFiiz

A

FApFiip

A

FFiip

FApFiF

FFiiF

_*

1__

*

1__

_

_*

1__

*

1__

_

1__

_

1__

_

_1_

_1_

1

1

1

1

1

1

Optimal operation point

the highest pressure difference of

booster

Behind the optimal point the pedal

force remains without boosting.Pressure of booster by optimal

point: pSt

Relative achievable deceleration

below optimal operation point

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Drum brakes

The brake pedal is actuated -> the piston pushes the brake shoes against the drum.

As the brake shoes contact the drum, there is a kind of wedging action, which has the effect of

pressing the shoes into the drum with more force.

The extra braking force provided by the wedging action allows drum brakes to use a smaller

piston than disc brakes. But, because of the wedging action, the shoes must be pulled away

from the drum when the brakes are released. This is the reason for some of the springs. Other

springs help hold the brake shoes in place and return the adjuster arm after it actuates.

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Drum brakes - overview

Connected shoes in twosenses (Duo servo)

One leading-trailingshoe (Simplex)

Two leading shoes(Duplex)

Brake with connected

shoes (Servo)

Two leading shoes in bothsense (Duo Duplex)

Source: Vlk F.

Podvozky motorovch vozidel

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Basic arrangements of drum brakes

Duo-Servo brake Two leading Shoe brake (Duplex)

Leading-Trailing Shoe Brake (Simplex)Source: Limpert R.


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Leading-Trailing Shoe Brake

Used as rear brake of passenger cars.

+ low sensitivity to lining friction

changes, stable brake production

Source: Limpert R.


Duo-Servo brake

The primary shoe reaction (at the bottomof the shoe) is used as application force

of the secondary shoe by pushing

through adjustement mechanism.

+ high brake torque

- high variation in brake torque for smallchanges of friction coefficient

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Drum brake

adjuster

For the drum brakes to function correctly, the brake shoes must remain close

to the drum without touching it. If they get too far away from the drum (as theshoes wear down, for instance), the piston will require more fluid to travel

that distance, and your brake pedal will sink closer to the floor when you

apply the brakes. This is why most drum brakes have anautomatic

adjuster.

As the pad wears down, more space will form between the shoe and thedrum. Each time the car stops while in reverse, the shoe is pulled tight

against the drum. When the gap gets big enough, the adjusting lever rocks

enough to advance the adjuster gear by one tooth. The adjuster has threads

on it, like a bolt, so that it unscrews a little bit when it turns, lengthening to fill

in the gap. When the brake shoes wear a little more, the adjuster can

advance again, so it always keeps the shoes close to the drum.

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Source: Limpert R.


S-cam with automatic slack adjuster

(Rockwell international)90 % of air brake trucks and

tractors use S-cam or wedge

actuated drum brake.

S-cam uses leading-trailing

shoe design. The shoes area

applied mechanically by

rotation of a cam shaped in anS form. Rotation of the cams

pushes the rollers and tips of

the shoes apart. Due to cam

geometry the application force

against the leading shoe willhave a smaller lever arm

relative to the pivot anchor of

the leading shoe than that of

the trailing shoe nearly

uniform wear of leading andtrailing shoe

leading shoe

Trailing shoe

S-cam

Air-brake chamber

Estimation of the drum brake torque

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Simplified theory

FL leading shoe tip resultant

FT Trailing shoe tip resultant

MB Brake drum torque

coefficient of friction

between lining and drumN radial force between lining

and drum

r drum radius

rNMB =

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Simplified calculationMoment balance around the point A

0

0

=+

=

=+

bFcFhF

FF

bFcFhF

dda

d

n

nda

Brake factor of the leading shoe

cb

h

F

FBF

a

d

leading

==

Total brake factor of the leading and trailing shoe

2

1

2

=

b

c

b

h

BF

Sensitivity

2

2

2

1

1

2

)(

+

==

b

c

b

cb

h

d

BFdS

Braking torque

aFrBFM =B

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Brake factor

BF1 Brake factor of the leading shoe

BF2 - Brake factor of the trailing shoe

BF Total brake factor

The curves are computed for thefollowing parameters of the brake:

h = 200 mm

b = 100 mm

c = 75 mm

Source: Limpert R.



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Real distribution of pressure

For calculation of real braking the magnitude and location of the resulting

force acting on each shoe should be determined.

leading shoe trailing shoe Example of the continous load distribution by

leading & trailing shoe drum brake.

Presumption:

The magnitude of the specific continous load onthe shoe pads corresponds to the magnitude of

the deformation.


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sin== ckkp

sinsinsin

sin

max==

=

pakpa

c

Deformation is proportional to the pressure

With usage of law of sinus in ASK

The pressure and deformation have sinusoidaldevelopment around the drum brake surface

The maximal pressure occurs in the location which

corresponds to the 90 deg measured from join line

between the drum center and anker point of the shoe.


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The magnitude and the location of the

resulting normal force we obtain by ploting

the line of elementary components of

normal force. When choosing small anglethe resulting curve will be cycloid.


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The angle of contact of brake pads is mostly smaller than 120 deg, and can

non symmetrically distributed round the drum circumference. When the

angle of contact equals21, the resultatiting normal force will be obtained

by linking the points on the cycloid.

The cycloid curves are similar one to each other. Therefor is sufficient to plot

one cycloid for all drum brakes.


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nt FF =

'

2

'

1)()()(

2

1

2

1

2

1

=== rddFdFrdrdFF tttt

rF

rF nt

== 21'

2

'

1

The frictional force will be obtain from the cycloid and the coefficient of

friction between lining and drum

The direction of the frictional force is perpendicula to the normal one.

The point of application can be obtain from the moment equilibrium to the

center point S

The fictive radius of application equals


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2211 += tt FFM

Remaining forces can be obtained graphically separately for leading and

trailing shoe. We choose the unit actuating force. The resulting braking

moment is sum of braking moments of both shoes.

Leading shoe Trailing shoe

Disc versus DrumSource: Limpert R.


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DiscversusDrum

brakesDisc brakes

+

little fade at high temperatures

no increase of pedal travellinear relationshipe between brake

torque and pad/disc friction coefficient

temperatures up to 1173K

Drum brakes

-

highly temperature sensitive

maximum temperature 700 K

the drum increases with temperatureincrease (by 1 to 1,5 mm)

larger drum diameter causes improper

contact between lining and drumBrake factor: ability of a brake to

produce brake torque for differentlining/drum friction coefficients

B k i

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Brake comparison

Very lowVery high5,0Duo-servo shoes

LowHigh3,0Two leading

shoes

ModerateModerate2,2Leading and

trailing shoes

LowHigh1,6Single leading

shoe

HighLow1,2Disc and pad

Very highVery low1,15Two trailing

shoes

Very highVery low0,55Single trailing

shoe

StabilityRelative

braking power

Brake factor

for~ 0,375

Type of brake

Comparison of

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Comparison of

brake system

control media

Safety factors

Only 2 % of highway accidents involvebrake malfunction as a contributing

accident causation. Of these 90 % are

related to the brake system defects

caused by improper maintentance.

Source: Limpert R.


Vehicle stability during braking

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Import experience for vehicle stability during braking:

By frontal crash with initial speed of 50 km/h in most cases the passengers

survive.By side crash with vehicle speed of 30 km/h in most cases the passengers

do not survive.

The vehicle should during braking always maintain its direction, even aftera disturbance should return to its previous direction.


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gxz &&= Braking ratio

zGZZ

zGBB

RRFF

RF

=+

=+

x:B

FBR

ZFZR

G

xm &&

lF lR

l

h

z

A

MA:

+=

=

l

hx

l

lgmZ

hxmlGlZ

RF

RF

&&

&& 0

z: GZZ RF =+

=

=

=

=

l

hx

l

lgmZ

l

hx

l

lgmZ

l

hxm

l

lgmgmZ

ZGZ

F

R

R

R

R

R

FR

&&

&&

&&

1

l

h

l

lF

=

=

( )

( )

=

+=

zGZ

zGZ

R

F

1


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( )

( )

( ) ( )

[ ]yBRyBFxBRxBF

z

yBR

F

yBF

F

xBR

F

xBF

Fz

yBRFyBFFxBRFxBFFz

FR

FF

yBRFyBFFxBRRxBFFz

qyBRyBF

xBRxBF

RF

FFFFJ

l

dt

d

Fl

llF

l

lF

l

llF

l

l

ldt

dJ

lFllFlFllFldt

dJM

llb

lb

FllFlFbFbdt

dJM

GzFFy

GzFFx

+=

+

=

+=

=

=

+=

=+

=+

=

==

cos)1(cossin)1(sin

coscossinsin

:/coscossinsin:

cos

sin

coscos:

:

:

2

2

2

2

2

2

2

2

Source: Burkhardt

Bremsdynamik und PkW Bremsanlagen


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Rigid

bicyclem

odel

( )

( )

cos

cos

1

=

=

=

+=

=

=

=

=

RhydxBR

FhydxBF

yBRyBR

yBFyBF

R

ybR

yBR

F

ybF

yBF

R

xbR

xBR

F

xbF

xBF

BpF

BpF

zGF

zGF

Z

F

Z

F

Z

F

Z

F

[ ]yBRyBFxBRxBF

z

FFFFJ

l

dt

d+=

cos)1(cossin)1(sin

2

2

Source: Burkhardt


Source: Burkhardt



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Rigid

bicyc

lemodel

[ ]yBRyBFxBRxBFz

FFFFJ

l

dt

d+=

cos)1(cossin)1(sin

2

2

Positivevalue of yaw rate increase of

body slip angle vehicle isinstable

Negativevalues decrease of body slip angle

vehicle isstable

In case offront axle blocking:FyBF= 0 => yaw rate negative => the vehicle remains

stable

Rear axle is blocked:

FyBR= 0 => yaw rate positive => immediate increaseof body slip angle => the vehicle isunstable

Source: Burkhardt


Stability of the vehicle during braking

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Stability of the vehicle during braking

85 % of braking torque

actuates the rear wheels

85 % of braking torque

actuates the front wheels

Source: Mitschke M. Wallentowitz H.



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BF

BRZF

ZR

G

xm &&

lF lR

l

h

y

A

+==+

==

==

=+=

G

Z

G

Zz

G

B

G

B

ZBZB

zG

Zz

G

Z

RFRF

RF

RRRFFF

RF

1

z=

Special case

G

BF

G

BR

Source: Burkhardt


Ideal force distribution during braking

In ideal case on the front and rear axle is thesame utilization of tangential forces


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Ideal force distribution during braking

( )

( )

=

+=

zzG

B

zzG

B

R

F 1

Elimination of z from equations leads to:G

B

G

B

G

B FFR +

=

2

11

4

)1(2

2

Equation of parabola Parabola of idealdistribution of tangential (braking) forces.

Parabol axis:

G

B

G

B RF ;

G

BR

G

BF

Source: Burkhardt


Analysis of parabola of ideal

b ki f di t ib ti

Source: Burkhardt


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G

BR

braking force distribution

G

BF

Case: no braking force on

rearaxle

1) z=0

2) -z=0

Case: no braking force on

frontaxle

1) z=0

2) 1-+z=0

Equation for the line of

parabola symmetry

=

4

21

G

B

G

B FR



Source: Burkhardt


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G

BR


G

BF

Coordinates of point A

Maximal value of BR/G is

achieved if

0=F

R

dB

dB

=+=

=

=

=

=+

2

4

2

4

14

)1(2

2

2

max

2

2

G

B

G

Bz

G

B

G

B

G

B

GB

FR

A

A

F

A

RR

F



Source: Burkhardt


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G

BR


G

BF

Coordinates of point B

Maximal value of BF/G is

achieved if

=F

R

dB

dB

The maximum is

negative, therefore

the result is valid

for traction, not forbraking

BFwas replaced

with FTFBRwas replaced

with FTR

( )

=+=

=

=

=

+

2

1

41

4

1

04

)1(

2

2

2

G

F

G

Fz

GF

G

F

GB

TRTF

B

TR

B

TF

F



Source: Burkhardt


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G

BR


G

BF

Parabola

symmetry

Slope of the parabola at any point:

1

4

)1(2

1

2

2

+

=

G

BdB

dB

FF

R

= 1FR

dB

dB

Slope of the parabola in the origin:

Diagram of braking force distribution

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g g

R

R

P

P

RF

critcrit

critkr

Z

B

Z

B

z

DBDB

z

===

=+

=

=

1

G

B R

G

BFSource: BurkhardtBremsdynamik und PkW Bremsanlagen

1 Parabola of ideal braking force distribution

2 Lines of installed brake distribution

3 Lines of front tire-road friction coefficients

4 Lines of rear tire-road friction coefficients

5 - Critical decelaration

Over

braked

fronta

xle

Overbra

ked

rearaxle

Limiting the braking force of rear

a le

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axle

1 Brake fluid input

2 Guidance pivots

3 Ball tube

4 Mass

5 Spring

6 Piston

7 Spring

9 Brake fluid output

10 - Sealing

The brake fluid of the circuit the rear axle comes from the mastercylinder to the input (1).In the limiter follows the hole of Guidance pivot (2). On (2) reside ball tube (3) and mass

(4). The spring (5) ensure that these parts can not move. The working point is

determined by the spring (5). For compensation of short impact are implemented piston

(6) and spring (7). The mass does not move till z=0,7. On the output (9) is connected

the rear axle circuit. When the working point is overcomed the mass moves and preventthe pressure increase in rear axle circuit

Source: Burkhardt


Limiter Bendix

Limiting the braking force of rear

axle

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axleSource: BurkhardtBremsdynamik und PkW Bremsanlagen

1 Control ball

2 Contact surface

3 Control opening

4 Brake fluid

5 Output

6 Stepped piston

ist control ratio

The limiter is in the vehicle mounted inclined with an angle. If the limit relativedecelarion overstep tan(), the ball (1) starts to roll to the control opening (3). So is the

fluid circulation to the output (5) blocked, i.e. the pressure in the rear axle can not

increase. If the input pressure is still increasing, after overstepping the second working

point the piston (6) moves. The pressure can again increase. The second working point

is determined by the ratio of input and output pressure corresponding to ist.

Limiter Girling

Function of limiters and ABS system

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y

Regulation depending

on statical load

Proportional reduction

valve controlABS Systems

Emergency braking without

any steering maneouver

1 Path of vehicle COG

2 Vehicle without limiter (ABS)

3 axleTwo-axle ABS with

separate wheel control

4 Two-axle ABS Select-Low

on the rear Source: Mitschke, WallentowitzDynamik der Kraftfahrzeuge

Braking of trucks and articulated

vehicles

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vehicles

The driver force is not sufficient to actuate thebrakes of truck. Therefor as the energy medium

is used compressed air. The brake pedal effortof the driver is used to modulate the pressureapplied to the brake chambers.

No manual push-through when the energy source

is off.Brake system must have a dual air brake system, if

one circuit fails, emergency braking function ismaintained.

Air-over-hydraulic brakes the air pressure isconverted into hydraulic pressure used to pressshoes against the drum.

Air brake system1 Air compressor

2 Compressor governor

3 Wet supply reservoir

4 Drain cock

5 S f t l

Source: Limpert R.


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The compressor1charges wet supply reservoir3 from which reservoirs9and13 arefed (and reservoirs of trailer). The dual brake system is modulated by the driver througthe dual brake application valve15. When brake application valve is released, all brakechambers16 exhaust their respective quick release valves (21). When front brakecircuite fails, double check valve12 and reservoir single check valve immediatelyclose. The same for rear brake failure. Because of double check valve, air is suppliedto the tractor and trailer spring brakes and the trailer service brakes if the tractor rearsystem becomes inoperative. If both the front and rear brake systems become

inoperative, spring brakes will apply automatically when the air pressure drops belowapproximately 40 psi.

5 Safety pressure valve

6 Pressure protection valve

7 Automatic drain valve8 One-way check valve

9 Front system reservoir

10 Low pressure switch

11 Automatic front brake

limiting valve (ratio valve)

12 Double check valve

13 Rear system reservoir14 Service relay valve (if

ABS wheel lock control

modulator)

15 Dual application valve

16 Service brake chamber

17 Spring brake chamber

18 System park controlvalve

19 Spring brake relay valve

20 Dual air gage

21 Quick release valve

22 Spring brake control

valve

23 Instrument package

manifold valve

24 Stoplight switch

25 Application pressure air

gage

26 Filler valve

Retarders

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Engine passive resistance in engine when thegear is engaged

Engine brake (exhaust brake) restriction ofexhaust gax output, change in valve timing,lowering the fuel supply (double or triple actionthan just braking by engine).

Hydrodynamic brake (need of cooling system,compact powerful system)

Electrodynamic brake (stator set ofelectromagnets linked with chassis, rotor is

driven by output shaft. Generated eddie currentproduces magnetic force which generatesbraking torque. (Simple design, high weight,dependence on temperature.)

Retarders

Source: Vlk F.

Podvozky motorocch vozidel

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Engine brake MAN EVB

1 Exhaust throttle

2 Restriction

3 Compressed air

4 - Piston

Edie current electrodynamic

brake:

1 Rotor

2 Stator

3 Locking caliper4 Driving shaft

Retarders - comparison

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Source: Vlk F.

Podvozky motorovch vozidel

Brakingtorque

Maximal

torqueofinternalcombustionengin

e

rpm

Rpm by nominal power

Eddie

current

brake

Hydrodynamicbrake

2 types of design

Brakingb

yICengin

eonly

ICengineequip

ped

withexha

ustbrake

Braking dynamics of

combination vehicles

Source: Vlk F.

Dynamika vozidel

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Black wheels are overbraked.

White wheels are braking.

The figure shows that cases e) and f)

are strongly unstable

Braking dynamics of


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Source: Vlk F.

Dynamika vozidel

Black wheels are overbraked.

White wheels are braking.

Documents

Brakes