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Erasmus LLP Intensive Programme

Your speaker

Mark Pecqueur

Onsiphore Pecqueur (1792 1852)

Professor at KdG University college

Independent expert on Engines and vehicles

More than 18 years active in the world of alternativefuels and drive systems

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Understeer & oversteer

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Understeer & oversteer

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Erasmus LLP Intensive Programme

The perfect steering system

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The perfect steering system

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If we work this out we get:

+

++

=

22

22

br

br

blrl

blrl

pQff

ff

Finaly:4

22

2 br

rlpQ

f

=

If we combine this with (d) we can calculate x:

4

2

2

2

2 br

rl

br

bl

x

f

f

f

=

r

bbr

r

bb

r

x

f

ff

f

+

=

+

=2

2

2

2

2

r

bbrx

ff

+=

4

22

This is the equation of the x

coordinate of the intersectionof straight AM and BS we cannow also calculate the ycoordinate in theintersection.

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The perfect steering system

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pxy =

Fill in x and P and we get:

2

2

2

b

r

l

r

bb

ry

f

f

f

+

+

=

In the end we get:

r

lby

f

=2

r

bbrx

ff

+=

4

22

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The perfect steering system

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24

2lim

2

b

r

bbrx

fff

r

=

+=

02

lim =

=

r

lby

f

r

02

,bf

In a first evaluation we calculate the

coordinates x and y during straightahead driving. This means that thecurvature is than equal to: Applyingthis to the equation we get:

During this situation the intersection is:

We find this point in the middle of thefront axle of the vehicle.

The second interesting point is the point with the smallest turning circle. This is when the vehicle isturning around his rear wheel the curvature is than equal to:

2

br

f= Aplying this value into the equation for x and y we

get:b

r

bbrx f

ff

rbf

=

+=

4

2lim

2

2

lr

lby

f

rbf

=

=

2

lim

2

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Different steering principles

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Turntable steering system

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E LLP I t i P

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Different steering principles

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Pivot steering system

0,00

5,00

10,00

15,00

20,00

25,00

30,00

5 7 911

13

15

17

19

21

23

25

27

29

31

33

35

37

39

41

43

45

47

49

51

53

55

wheelangle()

radius (m)

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Different steering principles

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Forces & Moment on the steeringangle center line.

Moments du to vertical forces

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Influence of or casterFncos()

Fnsin()

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Forces & Moment on the steeringangle center line.

Moments du to vertical forces

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M=Fnsin()sin()ra

offset tot increase stability

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Forces & Moment on the steeringangle center line.

Influence of the tire deformation on the moment around the king pin

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displacement from the camber offset while driving in a curve

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Properties of the steering system

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wa=(wl+wr)

is=ishiss

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Properties of the steering systemstiffness of the steering system

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g

Cornering performanceStationary cornering just not drifting out of the curve

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FFG

Fsrslc

+= )sin()cos(

( )FFGF nrnlc += )sin()cos(

+= )cos()sin()sin()cos( GFGF cc

rgvG

r

vmFc

=

=

22

+

=

)cos()sin()sin()cos(

22

Grg

vGG

rg

vG

The centriugal force is equal to:

+

=

1)()(

22

tgrg

vtgGrg

v

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g

Cornering performanceStationary cornering just not drifting put of the curve

)()(1

2

tgtgrg

v +=

Now we calculate the speed v:

)(1

)(2

tg

tg

rgv

+

=

rgtg

tgvg

+=

)(1

)(max

Let us suppose that we drive on a flat road we have than:

rgv =

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Cornering performanceCornering stability

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vv gk maxmax >

rgtg

tgrg

btgh

bhtg

+>

+

)(1

)(

2)(

2)(

Square both terms and deviding by gr:

+>

+

2

)()()(1

2

)(b

tghtgtgb

htg