Transcript Slide 1

Geometry and Linkage
Lecture 1
Day 1-Class 1
References
 Gillespie, T., The Fundamentals of Vehicle
Dynamics, Society of Automotive Engineers,
Warrendale, PA, 1992.
 Milliken, W.F. and Milliken, D.L., Chassis
Design Principles and Analysis, Society of
Automotive Engineers, Warrendale, PA, 2002.
 Hunt, D., Farm Power and Machinery
Management, Iowa State University Press,
Ames, IA, 2001.
Ackerman Geometry
δo
δi
 Basic layout for
passenger cars,
trucks, and ag
tractors
 δo = outer steering
angle and δi = inner
steering angle
 R= turn radius
 L= wheelbase and
t=distance between
tires
Center of
Gravity
L
Turn
Center
R
δi
t
δo
Figure 1.1.
Pivoting
Spindle
(Gillespie, 1992)
Cornering Stiffness and
Lateral Force of a Single Tire
 Lateral force (Fy) is the force produced
by the tire due to the slip angle.
 The cornering stiffness (Cα) is the rate of
change of the lateral force with the slip
α
angle.
C 
V
Fy

Fy
(1)
t
Figure 1.2. Fy
acts at a
distance (t) from
the wheel center
known as the
pneumatic trail
(Milliken, et. al., 2002)
Slip Angles
 The slip angle (α) is the angle at which a tire
rolls and is determined by the following
equations:
W f *V 2
f 
(2)
Cf * g * R
Wr *V 2
r 
Cr * g * R
(3)
Fy
α
V
t
W = weight on tires
C α= Cornering Stiffness
g = acceleration of gravity
Figure 1.2.
Repeated
V = vehicle velocity
(Gillespie, 1992)
Steering angle
 The steering angle (δ) is also known as the
Ackerman angle and is the average of the
front wheel angles
δo
 For low speeds it is:
δ
L
 
R
i
(4)
 For high speeds it is:
L
    f r
R
Center
of
L Gravity
R
(5)
δi
αf=front slip angle
αr=rear slip angle
t
δo
Figure 1.1.
Repeated
(Gillespie, 1992)
Three Wheel
Figure 1.3. Three wheel
vehicle with turn radius
and steering angle
shown
R
δ
 Easier to determine steer
angle
 Turn center is the intersection
of just two lines
Pivoting Single Axle
Figure 1.4. Pivoting
single axle with turn
radius and steering
angle shown
R
δ
 Entire axle steers
 Simple to determine steering angle
Both axles pivot
R
δ
Figure 1.5. Both axles
pivot with turn radius
and steering angle
shown
 Only two lines determine steering
angle and turning radius
 Can have a shorter turning radius
Articulated
 Can have
shorter turning
radius
 Allows front
and back axle
to be solid
Figure 1.6. Articulated
vehicle with turn radius
and steering angle
shown
Aligning Torque of a Single
Tire
 Aligning Torque (Mz) is the resultant
moment about the center of the wheel do
to the lateral force.
M z  Fy * t
Figure 1.7. Top
view of a tire
showing the
aligning torque.
α
(6)
V
Fy
t
Mz
(Milliken, et. al., 2002)
Camber Angle
 Camber angle (Φ) is
the angle between
the wheel center and
the vertical.
 It can also be
referred to as
inclination angle (γ).
Φ
Figure 1.8.
Camber angle
(Milliken, et. al., 2002)
Camber Thrust
 Camber thrust
(FYc) is due to the
wheel rolling at
the camber angle
 The thrust occurs
at small distance
(tc) from the
wheel center
 A camber torque
is then produced
(MZc)
Mzc
tc
Fyc
Figure 1.9. Camber thrust and torque
(Milliken, et. al., 2002)
Camber on Ag Tractor
Pivot Axis
Φ
Figure 1.10.
Camber angle on
an actual tractor
Wheel Caster
 The axle is placed
Pivot Axis
some distance
behind the pivot axis
 Promotes stability
 Steering becomes
more difficult
Figure 1.11. Wheel
caster creating
stability
(Milliken, et. al., 2002)
Neutral Steer
 No change in the steer angle is
necessary as speed changes
 The steer angle will then be equal to the
Ackerman angle.
 Front and rear slip angles are equal
(Gillespie, 1992)
Understeer
 The steered wheels must be steered
to a greater angle than the rear
wheels
 The steer angle on a constant radius
turn is increased by the understeer
gradient (K) times the lateral
acceleration.
L
   K * ay
R
(7)
ay
α
V
t
Figure 1.2.
Repeated
(Gillespie, 1992)
Understeer Gradient
 If we set equation 6 equal to equation 2
we can see that K*ay is equal to the
difference in front and rear slip angles.
 Substituting equations 3 and 4 in for the
slip angles yields:
K
Wf
Cf
Wr

Cr
(8)
Since
2
V
ay 
g*R
(9)
(Gillespie, 1992)
Characteristic Speed
 The characteristic speed is a way to
quantify understeer.
 Speed at which the steer angle is
twice the Ackerman angle.
Vchar
57.3 * L * g

K
(10)
(Gillespie, 1992)
Oversteer
 The vehicle is such that the
steering wheel must be turned so
that the steering angle decreases
as speed is increased
 The steering angle is decreased
by the understeer gradient times
the lateral acceleration, meaning
the understeer gradient is
negative
 Front steer angle is less than rear
steer angle
(Gillespie, 1992)
Critical Speed
 The critical speed is the speed
where an oversteer vehicle is no
longer directionally stable.
Vcrit
57.3 * L * g

K
(11)
Note: K is negative in oversteer case
(Gillespie, 1992)
Lateral Acceleration Gain
 Lateral acceleration gain is the ratio
of lateral acceleration to the steering
angle.
 Helps to quantify the performance of
the system by telling us how much
lateral acceleration is achieved per
degree of steer angle
V2
ay
57.3Lg

2
KV

1
57.3Lg
(12)
(Gillespie, 1992)
Example Problem
 A car has a weight of 1850 lb front axle and
1550 lb on the rear with a wheelbase of
105 inches. The tires have the cornering
stiffness values given below:
Load
lb/tire
Cornering
Stiffness
lbs/deg
Cornering
Coefficient
lb/lb/deg
225
74
0.284
425
115
0.272
625
156
0.260
925
218
0.242
1125
260
0.230
Determine the steer angle if the
minimum turn radius is 75 ft
 We just use equation 1.
L 105 / 12
 
 0.117
R
75
Or 6.68 deg
Find the Understeer gradient
 The load on each front tire is 925 lbs and the
load on each rear tire is 775 lbs
 The front cornering stiffness is 218 lb/deg and
the rear cornering stiffness 187 lb/deg (by
interpolation)
 Using equation 7:
K 
Wf
Cf
Wr

Cr
925lb
775lb


218lb / deg 187lb / deg
 0.099deg(/ g )
Find the characteristic speed
 Use equation 8 plugging in the given
wheelbase and the understeer gradient
Vchar
57.3 * L * g

K
57.3 deg/ rad *105in * 32.2 ft / s 2

12in / ft * 0.099deg
 404 ft / s
 275m ph
Determine the lateral acceleration
gain if velocity is 55 mph
 Use equation 10
V2
ay
57.3Lg

KV 2

1
57.3Lg
(81 ft / s ) 2
57.3 deg/ rad (105in / 12in / ft )(32.2 ft / s )

0.099deg/ g (81 ft / s ) 2
1
57.3 deg/ rad (105in / 12in / ft )(32.2 ft / s )
 0.391g / deg