Transcript
Vehicle Load Transfer
Wm Harbin
Technical Director
BND TechSource
1
Vehicle Load Transfer
Part I
General Load Transfer
2
Factors in Vehicle Dynamics
Within any modern vehicle suspension there are many factors
to consider during design and development.
Factors in vehicle dynamics:
• Vehicle Configuration
• Vehicle Type (i.e. 2 dr Coupe, 4dr Sedan, Minivan, Truck, etc.)
• Vehicle Architecture (i.e. FWD vs. RWD, 2WD vs.4WD, etc.)
• Chassis Architecture (i.e. type: tubular, monocoque, etc. ; material: steel, aluminum,
carbon fiber, etc. ; fabrication: welding, stamping, forming, etc.)
• Front Suspension System Type (i.e. MacPherson strut, SLA Double Wishbone, etc.)
• Type of Steering Actuator (i.e. Rack and Pinion vs. Recirculating Ball)
• Type of Braking System (i.e. Disc (front & rear) vs. Disc (front) & Drum (rear))
• Rear Suspension System Type (i.e. Beam Axle, Multi-link, Solid Axle, etc.)
• Suspension/Braking Control Systems (i.e. ABS, Electronic Stability Control,
Electronic Damping Control, etc.)
3
Factors in Vehicle Dynamics
Factors in vehicle dynamics (continued):
• Vehicle Suspension Geometry
• Vehicle Wheelbase
• Vehicle Track Width Front and Rear
• Wheels and Tires
• Vehicle Weight and Distribution
• Vehicle Center of Gravity
• Sprung and Unsprung Weight
• Springs Motion Ratio
• Chassis Ride Height and Static Deflection
• Turning Circle or Turning Radius (Ackermann Steering Geometry)
• Suspension Jounce and Rebound
• Vehicle Suspension Hard Points:
• Front Suspension
• Scrub (Pivot) Radius
• Steering (Kingpin) Inclination Angle (SAI)
• Caster Angle
• Mechanical (or caster) trail
• Toe Angle
• Camber Angle
• Ball Joint Pivot Points
• Control Arm Chassis Attachment Points
• Knuckle/Brakes/Steering
• Springs/Shock Absorbers/Struts
• ARB (anti-roll bar)
4
Factors in Vehicle Dynamics
Factors in vehicle dynamics (continued):
• Vehicle Suspension Geometry (continued)
•Vehicle Suspension Hard Points (continued):
• Rear Suspension
• Scrub (Pivot) Radius
• Steering (Kingpin) Inclination Angle (SAI)
• Caster Angle (if applicable)
• Mechanical (or caster) trail (if applicable)
• Toe Angle
• Camber Angle
• Knuckle and Chassis Attachment Points
• Various links and arms depend upon the Rear Suspension configuration.
(i.e. Dependent vs. Semi-Independent vs. Independent Suspension)
• Knuckle/Brakes
• Springs/Shock Absorbers
• ARB (anti-roll bar)
•Vehicle Dynamic Considerations
• Suspension Dynamic Targets
• Wheel Frequency
• Bushing Compliance
• Lateral Load Transfer with and w/o ARB
• Roll moment
• Roll Stiffness (degrees per g of lateral acceleration)
• Maximum Steady State lateral acceleration (in understeer mode)
• Rollover Threshold (lateral g load)
• Linear Range Understeer (typically between 0g and 0.4g)
5
Factors in Vehicle Dynamics
Factors in vehicle dynamics (continued):
• Vehicle Dynamic Considerations (continued)
• Suspension Dynamic Analysis
• Bundorf Analysis
• Slip angles (degrees per lateral force)
• Tire Cornering Coefficient (lateral force as a percent of rated vertical load per degree slip
angle)
• Tire Cornering Forces (lateral cornering force as a function of slip angle)
• Linear Range Understeer
• Steering Analysis
• Bump Steer Analysis
• Roll Steer Analysis
• Tractive Force Steer Analysis
• Brake Force Steer Analysis
• Ackerman change with steering angle
• Roll Analysis
• Camber gain in roll (front & rear)
• Caster gain in roll (front & rear – if applicable)
• Roll Axis Analysis
• Roll Center Height Analysis
• Instantaneous Center Analysis
• Track Analysis
• Load Transfer Analysis
• Unsprung and Sprung weight transfer
• Jacking Forces
• Roll Couple Percentage Analysis
• Total Lateral Load Transfer Distribution (TLLTD)
6
Vehicle Load Transfer
While the total amount of factors may seem a bit overwhelming, it may be
easier to digest if we break it down into certain aspects of the total.
The intent of this document is to give the reader a better understanding of
vehicle dynamic longitudinal and lateral load transfer as a vehicle
accelerates/decelerates in a particular direction.
The discussion will include:
Part I – General Load Transfer Information
•
•
•
Load vs. Weight Transfer
Rotational Moments of Inertia
Sprung and Unsprung Weight
Part III – Lateral Load Transfer
•
•
•
•
Part II – Longitudinal Load Transfer
•
•
•
Vehicle Center of Gravity
Longitudinal Load Transfer
Suspension Geometry
•
•
•
Instant Centers
Side View Swing Arm
Anti-squat, Anti-dive, and Anti-lift
Cornering Forces
Suspension Geometry
• Front View Swing Arm
•
Roll Stiffness
• Anti-roll bars
•
•
•
Roll Center Heights
Roll Axis
Tire Rates
Roll Gradient
Lateral Load Transfer
7
Load vs. Weight Transfer
8
Load vs. Weight Transfer
In automobiles, load transfer is the imaginary "shifting" of
weight around a motor vehicle during acceleration (both
longitudinal and lateral). This includes braking, or deceleration
(which can be viewed as acceleration at a negative rate). Load
transfer is a crucial concept in understanding vehicle dynamics.
Often load transfer is misguidedly referred to as weight
transfer due to their close relationship. The difference being
load transfer is an imaginary shift in weight due to an
imbalance of forces, while weight transfer involves the actual
movement of the vehicles center of gravity (Cg). Both result in
a redistribution of the total vehicle load between the
individual tires.
9
Load vs. Weight Transfer
Weight transfer involves the actual (small) movement of the
vehicle Cg relative to the wheel axes due to displacement of
liquids within the vehicle, which results in a redistribution of
the total vehicle load between the individual tires.
Liquids, such as fuel, readily flow within their containers,
causing changes in the vehicle's Cg. As fuel is consumed, not
only does the position of the Cg change, but the total weight
of the vehicle is also reduced.
Another factor that changes the vehicle’s Cg is the expansion
of the tires during rotation. This is called “dynamic rolling
radius” and is effected by wheel-speed, temperature, inflation
pressure, tire compound, and tire construction. It raises the
vehicle’s Cg slightly as the wheel-speed increases.
10
Load vs. Weight Transfer
The major forces that accelerate a vehicle occur at the tires
contact patch. Since these forces are not directed through the
vehicle's Cg, one or more moments are generated. It is these
moments that cause variation in the load distributed between
the tires.
Lowering the Cg towards the ground is one method of
reducing load transfer. As a result load transfer is reduced in
both the longitudinal and lateral directions. Another method
of reducing load transfer is by increasing the wheel spacings.
Increasing the vehicles wheel base (length) reduces
longitudinal load transfer. While increasing the vehicles track
(width) reduces lateral load transfer.
11
Rotational Moments of Inertia
Cg
Yaw
(r)
Roll
(p)
Pitch
(q)
y
Lateral
x
Longitudinal
z
Vertical
12
Moment of Inertia
Polar moment of inertia
• A simple demonstration of polar moment of inertia is to compare a
dumbbell vs. a barbell both at the same weight. Hold each in the
middle and twist to feel the force reacting at the center. Notice the
dumbbell (which has a lower polar moment) reacts quickly and the
barbell (which has a higher polar moment) reacts slowly.
o Wd
d1
d1
Example:
W = 50 lb (25 lb
at each end)
d1 = 8 in
d2 = 30 in
CL
CL
2
d2
d2
W
1 2 * 25 * (8)2 3200lb in2
W
2 2 * 25 * (30)2 45,000lb in2
13
Moment of Inertia
Sum the polar moments of inertia
• The total polar moment of inertia for a vehicle can be
determined by multiplying the weight of each component by
the distance from the component Cg to the Cg of the vehicle.
The sum of the component polar moments of inertia would
establish the total vehicle polar moment of inertia.
• A vehicle with most of its weight near the vehicle Cg has a
lower total polar moment of inertia is quicker to respond to
steering inputs.
• A vehicle with a high polar moment is slower to react to
steering inputs and is therefore more stable at high speed
straight line driving.
14
Moment of Inertia
Effects of polar moments of inertia
• Here is an example of a V8 engine with a typical transmission
packaged into a sports car.
M(
M(
Example:
WEng = 600 lb
WTran = 240 lb
dEng = 40 in
dTran = 10 in
o
) WEng(dEng)2 WTran(dTran)2
o
) 600lb(40in)2 240lb(10in)2 984,000lb in2
dEng
dTran
15
Moment of Inertia
Effects of polar moments of inertia
• Here is an example of a V8 engine with a typical transmission
packaged into a sedan.
2
2
M
(
)
W
Eng
(
d
Eng
)
W
Tran
(
d
Tran
)
o
2
2
2
M
(
)
600
lb
(
70
in
)
240
lb
(
40
in
)
3
,
324
,
000
lb
in
o
Example:
WEng = 600 lb
WTran = 240 lb
dEng = 70 in
dTran = 40 in
dEng
dTran
16
Load Transfer
17
Load Transfer
Load Transfer
• The forces that enable a road vehicle to accelerate and stop
all act at the road surface.
• The center of gravity, which is located considerably above
the road surface, and which is acted upon by the
accelerations resulting from the longitudinal forces at the
tire patches, generates a moment which transfers load.
• As asymmetric load results in differing traction limits, a
vehicles handling is affected by the “dynamic load
distribution”.
18
Load Transfer
Load Transfer equations & terms
Newton' s Second Law : F = m a
Vehicle Weight *Vehicle Acceleration
Inertial Force
g
Inertial Force *CGheight
Load Transfer
Wheelbase
F = force
m = mass
a = acceleration
g = ag = acceleration due to gravity
= 32.2ft/sec2 = 9.8m/sec2
ax = acceleration in the x direction
ay = acceleration in the y direction
az = acceleration in the z direction
Weight = mass * ag
19
Load Transfer
Load transfers between the Center of Gravity and the road
surface through a variety of paths.
• Suspension Geometry
• Front: Location of instant centers
(Side View Swing Arm)
• Rear: Instant centers, Lift Bars
(Side View Swing Arm)
• Suspension Springs
• Front: Coils, Air Springs, leafs or Torsion bars and Antiroll bars
• Rear: Coils, Air Springs, leafs or Torsion bars and Antiroll bars
20
Load Transfer
Load transfer (continued)
• Dampers (Shock Absorbers)
• During transient conditions
• Tires
• During all conditions (where the rubber meets the road)
Where and how you balance the load transfer between the
Springs, Geometry, Dampers and Tires are key determinates as
to how well the car will accelerate and brake and the stability
associated with each condition.
21
Load Transfer Control Devices
Dampers (Shock Absorbers)
• Along with the springs, dampers transfer the load of the rolling
(pitching) component of the vehicle. They determine how the
load is transferred to and from the individual wheels while the
chassis is rolling and/or pitching.
• Within 65-70% critically damped is said to be the ideal damper
setting for both handling and comfort simultaneously. Most
modern dampers show some digression to them as well,
meaning they may be 70% critically damped at low piston
speeds but move lower to allow the absorption of large bumps.
Damping is most important below 4 in/second as this is where
car control tuning takes place.
22
Load Transfer Control Devices
Springs
• Along with the dampers (shock absorbers), springs transfer the load
of the sprung mass of the car to the road surface. During
maneuvers, depending on instant center locations, the springs and
dampers transfer some portion of the (m x a), mass x acceleration,
forces to the ground.
• Spring Rate is force per unit displacement for a suspension spring
alone .
• For coil springs this is measured axially along the centerline.
• For torsion bar springs it is measured at the attachment arm.
• For leaf springs it is measured at the axle seat.
• The spring rate may be linear (force increases proportionally with
displacement) or nonlinear (increasing or decreasing rate with
increasing displacement).
• Units are typically lb/in.
23
Load Transfer Control Devices
Anti-roll bars
• [Drawing 1] shows how an anti-roll bar (ARB) is twisted when
the body rolls in a turn. This creates forces at the four points
where the bar is attached to the vehicle. The forces are shown
in [Drawing 2]. Forces A on the suspension increase [load]
transfer to the outside tire. Forces B on the frame resist body
roll. The effect is a reduction of body roll and an increase in
[load] transfer at the end of the chassis which has the anti-roll
bar. Because the total [load] transfer due to centripetal force is
not changed, the opposite end of the chassis has reduced [load]
transfer. [6]
B
B
A
A
Drawing 2
Direction of Turn
Drawing 1
24
Load Transfer Control Devices
Bushing Deflection (suspension compliance)
• All of the calculations shown in this presentation do not include
bushing deflection. There are many rubber bushings within a vehicle
suspension to consider when analyzing suspension compliance.
25
Load Transfer Control Devices
Frame/Chassis Deflection
• All of the calculations shown in this presentation are made under
the assumption that the frame or chassis is completely rigid (both in
torsion and bending). Of course any flexing within the frame/chassis
will adversely effect the performance of the suspension which is
attached to it.
26
Sprung and Unsprung Weight
100% Unsprung weight includes the mass of the tires, rims,
brake rotors, brake calipers, knuckle assemblies, and ball
joints which move in unison with the wheels.
50% unsprung and 50% sprung weight would be comprised
of the linkages of the wheel assembly to the chassis.
The % unsprung weight of the shocks, springs and anti-roll
bar ends would be a function of their motion ratio/2 with
the remainder as % sprung weight.
The rest of the mass is on the vehicle side of the springs is
suspended and is 100% sprung weight.
27
Sprung and Unsprung Weight
28
Springs Motion Ratio
The shocks, springs, struts and anti-roll bars are normally mounted at
some angle from the suspension to the chassis.
Motion Ratio: If you were to move the wheel 1 inch and the spring were
to deflect 0.75 inches then the motion ratio would be 0.75 in/in.
Motion Ratio = (B/A) * sin(spring angle)
29
Springs Motion Ratio
The shocks, springs, struts and anti-roll bars are normally mounted at
some angle from the suspension to the chassis.
Motion Ratio: If you were to move the wheel 1 inch and the spring were
to deflect 0.75 inches then the motion ratio would be 0.75 in/in.
Motion Ratio =
(B/A) * sin(spring angle)
30
Wheel Rates
Wheel Rates are calculated by taking the square of the motion ratio times
the spring rate. Squaring the ratio is because the ratio has two effects on
the wheel rate. The ratio applies to both the force and distance traveled.
Because it's a force, and the lever arm is multiplied twice.
• The motion ratio is factored once to account for the distance-traveled differential
of the two points (A and B in the example below).
• Then the motion ratio is factored again to account for the lever-arm force
differential.
Wheel Rate (lb/in) =
Example:
K
(Motion Ratio)2* (Spring Rate)
|
A----B------------P
• P is the pivot point, B is the spring mount, and A is the wheel. Here the motion
ration (MR) is 0.75... imagine a spring K that is rated at 100 lb/in placed at B
perpendicular to the line AP. If you want to move A 1 in vertically upward, B would
only move (1in)(MR) = 0.75 in. Since K is 100 lb/in, and B has only moved 0.75 in,
there's a force at B of 75 lb. If you balance the moments about P, you get
75(B)=X(A), and we know B = 0.75A, so you get 75(0.75A) = X(A). A's cancel and you
get X=75(0.75)=56.25. Which is [100(MR)](MR) or 100(MR)2.
31
Wheel Rates
Since the linkages pivot, the spring angles change as the components
swing along an arc path. This causes the motion ratio to be calculated
through a range. The graph below shows an example of these results for
both coil-over shock and anti-roll bar for an independent front suspension
from rebound to jounce positions.
Example:
Coil-over KS = 400 lb/in (linear)
Coil-over MR = 0.72-.079 in/in
ARB KS = 451.8 lb/in (body roll)
ARB MR = 0.56-0.61 in/in
300
KW = Wheel Rate (lb/in) =
(Motion Ratio - range)2* (Spring Rate - linear)
KW = MR2 * KS
Wheel Rate vs. Wheel Position
250
Wheel Rate (lb/in)
200
Ride height
150
100
Coil-over Shock
ARB
50
0
-2.49 -2.22 -1.95 -1.68 -1.42 -1.15 -0.89 -0.63 -0.36 -0.10 0.16 0.41 0.67 0.93 1.18 1.44 1.69 1.94 2.19 2.44
Rebound to Jounce (in)
32
Wheel Rates
In longitudinal pitch, the anti-roll bar (ARB) rotates evenly as the chassis
moves relative to the suspension. Therefore, the ARB only comes into play
during lateral pitch (body roll) of the vehicle (it also comes into play during
one wheel bump, but that rate is not shown here).
Example:
Coil-over KS = 400 lb/in (linear)
Coil-over MR = 0.72-.079 in/in
ARB KS = 451.8 lb/in (body roll)
ARB MR = 0.56-0.61 in/in
300
KW = Wheel Rate (lb/in) =
(Motion Ratio - range)2* (Spring Rate - linear)
KW = MR2 * KS
Wheel Rate vs. Wheel Position
250
Wheel Rate (lb/in)
200
Ride height
150
100
Coil-over Shock
ARB
50
0
-2.49 -2.22 -1.95 -1.68 -1.42 -1.15 -0.89 -0.63 -0.36 -0.10 0.16 0.41 0.67 0.93 1.18 1.44 1.69 1.94 2.19 2.44
Rebound to Jounce (in)
33
Spring Rates/Ride Frequency
The static deflection of the suspension determines its natural frequency.
Static deflection is the rate at which the suspension compresses in
response to weight.
Ride Natural Frequency vs. Static Wheel Deflection
200
180
160
w = Frequency (cycles/min)
140
120
188
x
100
80
60
40
20
0
0
10
20
30
40
50
60
x = Static Deflection (in)
70
80
90
100
34
Spring Rates/Ride Frequency
Ride frequency is the undamped natural frequency of the body in ride.
The higher the frequency, the stiffer the ride.
Based on the application, there are ballpark numbers to consider.
•
•
•
30 - 70 CPM for passenger cars
70 - 120 CPM for high-performance sports cars
120 - 300+ CPM for high downforce race cars
It is common to run a spring frequency higher in the rear than the front.
The idea is to have the oscillation of the front suspension finish at the
same time as the rear.
Since the delay between when the front suspension hits a bump and the
rear suspension hits that bump varies according to vehicle speed, the
spring frequency increase in the rear also varies according to the
particular speed one wants to optimize for.
35
Spring Rates/Ride Frequency
Once the motion ratios has been established, the front and rear spring
rates can be optimized for a “flat” ride at a particular speed.
36
Spring Rates/Ride Frequency
Here are the equations from the previous spreadsheet:
KW K S * (MR)2 Spring EffectWheel Rate
KR
K T * KW
Ride Rate
K T KW
DS
WS
Static Deflection
KR
188
Natural Frequency (CPM)
DS
Rrec
K RR rec
Where:
KS = Spring Rate (lb/in)
MR = Motion Ratio (in/in)
KT = Tire Stiffness Rate (lb/in)
WS = Sprung Weight (lb)
w/60 = Hz
wf = Front Frequency (Hz)
l = Wheelbase (in)
v = Vehicle Speed (mph)
1 mph = 17.6 in/sec
1
Rec. Rear Natural Frequency (Hz)
(1 / f ) (1 / 17.6) * ( / v)
4 2 * Rrec 2 *WS
Rec. Rear Ride Rate
(12 * 32.2)
K SR rec
(KTrear * K RRrec ) /(KTrear K RRrec )
Rec. Rear Spring Rate
2
(MRrear )
37
Ride Rate
Ride Rate (independent suspension)
• The overall ride rate for a suspension can be thought of as a
series combination of two springs.
1.
2.
The wheel center rate acting between the chassis and the
wheel center.
The vertical tire rate acting between the wheel center and the
ground.
K T * KW
KR
K T KW
Where:
KR = overall ride rate (lb-in)
KT = vertical tire rate (lb/in)
KW = spring wheel rate (lb/in)
(independent suspension)
38
Ride Rate
Ride Rate (solid axle)
• The overall ride rate for a suspension can be thought of as a
series combination of two springs.
1.
2.
The wheel center rate acting between the chassis and the
wheel center.
The vertical tire rate acting between the wheel center and the
ground.
K T * KW
KR
K T KW
Where:
KR= overall ride rate (lb-in)
KT = vertical tire rate (lb/in)
KW = vertical axle rate (lb/in)
(solid axle suspension)
39
Vehicle Load Transfer
Part III
Lateral Load Transfer
40
Lateral Load Transfer
Lateral Load Transfer
• Now that the ride rate analysis is complete, we can move on to the
roll analysis. We will want to calculate the anti-roll bars. To do this we
will need the following information on the vehicle suspension:
•
•
•
•
•
•
•
•
•
•
•
•
•
•
Roll center heights front and rear
Roll Axis
Tire Static Load Radius
Tire Stiffness Rate
Spring motion ratio
ARB motion ratio
Track width (independent susp)
Leaf spring spacing (solid axle)
Sprung mass CG height
Sprung mass weight distribution
Roll Moment lever arm
Roll Moment per lateral g acceleration
Roll Stiffness Rate per Roll Gradient
Total Lateral Load Transfer Distribution (TLLTD)
41
Roll Centers and Roll Axis
Roll Centers and Roll Axis
• As the vehicle changes direction, the sprung mass (body) of a
vehicle pivots about its roll axis in the opposite direction. This
lateral load transfer is a result of the centripetal force acting
on the moment (distance) between the roll axis and the CG of
the vehicle.
• How the roll centers react to the suspension dynamics is called
the roll center characteristics. The roll center characteristics
affects the roll center height as well as camber changes caused
by movement of the roll center throughout the suspension
travel.
• There are many types of vehicle suspension designs; each has
unique roll center characteristics.
42
Roll Centers
Roll Centers
• Every vehicle has two roll centers, one at the front and one at
the rear. Each roll center is located at the intersection of a line
drawn from the center of the tire contact patch through the IC
of that tire’s suspension geometry.
• As the IC moves during suspension travel, so too will the roll
center.
43
Roll Centers
Roll Centers
• This example shows a solid rear axle with leaf springs. The
axle/differential is suspended and moves with the wheel
assemblies. The roll center height (Zrc) is derived by
intersecting a plane running through the spring pivots with a
vertical plane running through the centerline of the axle. The
roll center point is equal distance between the springs.
Roll Center
Zrc
Zrc = 16 in
44
Roll Centers
Roll Centers
• This example shows an independent rear suspension with the
differential attached to the chassis and control arms
suspending the wheel assemblies from the chassis. The control
arms are parallel, therefore the IC is infinite. In this case the
lines running from the center of the tire contact path to the
roll center are parallel with the control arms.
Control Arms
Zr = 0.896 in
Roll Center
Zr
45
Roll Centers
Roll Centers
• This example shows an independent front suspension with the
control arms suspending the wheel assemblies from the
chassis. Lines running through the control arm pivots and ball
joints intersect at the IC. The lines running from the center of
the tire contact path to the IC intersect at the roll center.
Control Arms
Ball Joints
Ball Joints
To IC
Zf = 0.497 in
To IC
Roll Center
Zf
46
Roll Centers and Roll Axis
Roll Axis
• A line connecting the two roll centers is called the roll axis.
Rear Roll Center
Vehicle Center of Gravity
Roll Axis
Front Roll Center
47
Tire Stiffness Rate
Tire Rate
• There are many ways to calculate tire rate.
•
•
•
•
•
Load-deflection (LD)
Non-rolling vertical free vibration (NR-FV)
Non-rolling equilibrium load-deflection (NR-LD)
Rolling vertical free vibration (R-FV)
Rolling equilibrium load-deflection (R-LD)
• The simplest would be load deflection.
• All tire manufacturers list a static load radius in their catalog for
a specific tire. They will also list that tire’s unloaded diameter.
There will also be a chart showing the tire’s maximum load
rating. From these numbers the static deflection can easily be
calculated and the static rate is load/deflection.
48
Tire Stiffness Rate
Tire Rate
•
•
These examples are for the static rate of the tires shown.
The tire stiffness rate will change due to changes in air pressure (D1 bar can
affect rate by 40%) and slightly (less than 1%) when the vehicle speed changes.
2039.3 lb
kt
1411.3 lb / in
(28.95in / 2) 13.03in
Example 1:
Tire = P 235/70 R 16 105 T
Static (max) load radius = 330.9mm (13.03in)
Unloaded diameter = 735.4mm (28.95in)
Max Load = 925kg (2039.3lb) at 2.5 bar
1521.2 lb
kt
1560.2 lb / in
(25.69in / 2) 11.87in
Example 2:
Tire = P 245/45 R 17 95 W
Static (max) load radius = 301.5mm (11.87in)
Unloaded diameter = 652.4mm (25.69in)
Max Load = 690kg (1521.2lb) at 2.5 bar
1708.6 lb
kt
1761.4 lb / in
(26.66in / 2) 12.36in
Example 3:
Tire = P 275/40 R 18 99 W
Static (max) load radius = 314mm (12.36in)
Unloaded diameter = 677.2mm (26.66in)
Max Load = 775kg (1708.6lb) at 2.5 bar
Source: BND TechSource Tire Data Calculator
49
Tire Static Load Radius
Tire Static Load Radius
•
•
The term Static Load Radius used to determine the tire stiffness rate is given
by the tire manufacturer at the maximum load value for that particular tire.
The Tire Loaded Radius (RLF;RLR) used in the following equations is calculated
at the vehicle corner load values (deflection=load/rate).
RL (28.95 / 2) (992 / 1411.3) 13.77in
Example 1:
Tire = P 235/70 R 16 105 T
Unloaded diameter = 735.4mm (28.95in)
Veh Corner Load = 450kg (992lb) at 2.5 bar
KT = 1382.6lb/in
RLF (25.69 / 2) (893 / 1560.2) 12.27in
Example 2:
Tire = P 245/45 R 17 95 W
Unloaded diameter = 652.4mm (25.69in)
Veh Corner Load = 405kg (893lb) at 2.5 bar
KT = 1560.2lb/in
RLR (26.66 / 2) (882 / 1761.4) 12.83in
Example 3:
Tire = P 275/40 R 18 99 W
Unloaded diameter = 677.2mm (26.66in)
Veh Corner Load = 400kg (882lb) at 2.5 bar
KT = 1761.4lb/in
Source: BND TechSource Tire Data Calculator
50
Sprung and Unsprung Weight
Sprung and Unsprung Weight
• An example of this would be the front unsprung weight is
11.5% (split equally left to right) of the vehicle weight. The rear
unsprung weight is 13.5% (split equally left to right) and then
the body would make up the remainder as sprung weight at
75%.
Right Front
Unsprung
Cg
Right Rear
Unsprung
Cg
Body
Sprung Cg
Left Rear
Unsprung
Cg
Left Front
Unsprung
Cg
51
Sprung and Unsprung Weight
Sprung and Unsprung Weight
• The sprung weight of the vehicle is simply the total weight minus the
unsprung weight.
Ws W WU1 WU2 WU3 WU4 Sprung Weight
l
a
WU2
b
WU4
x1
x1
y’
ys’
x
W
ys’’
Ws
WU1
as
tr
y’’
bs
WU3
52
Sprung and Unsprung Weight
Sprung and Unsprung Weight
• The sprung weight of the vehicle is simply the total weight minus the
unsprung weight.
Ws 3542 105 105 118.5 118.5
Example C3 Corvette Upgrade:
WS = Sprung weight (lb)
WT = 3542lb
WU1 = 105lb
WU2 = 105lb
WU3 = 118.5lb
WU4 = 118.5lb
W
WS 3095lb
l
a
b
U2
WU4
x1
x1
y’
ys’
x
W
ys’’
Ws
WU1
as
tr
y’’
bs
WU3
53
Sprung Weight
Sprung Weight Distribution
• The unsprung weight front and rear:
WUF WU1 WU2 Unsprung Weight Front
WUR WU 3 WU 4 Unsprung Weight Rear
l
a
WU2
b
WU4
x1
x1
y’
ys’
x
W
ys’’
Ws
WU1
as
tr
y’’
bs
WU3
54
Sprung Weight
Sprung Weight Distribution
• The unsprung weight front and rear:
WUF 105 105 210lb
Example C3 Corvette Upgrade:
WUF = Sprung weight front (lb)
WUR = Sprung weight rear (lb)
WU1 = 105lb
WU2 = 105lb
WU3 = 118.5lb
WU4 = 118.5lb
WUR 118.5 118.5 237lb
l
a
WU2
b
WU4
x1
x1
y’
ys’
x
W
ys’’
Ws
WU1
as
tr
y’’
bs
WU3
55
Sprung Weight
Sprung Weight Distribution
• Taking the moments about the rear axle gives the longitudinal
location of the sprung mass CG.
bs
(WT * b) (WUF * )
Ws
as bs
and
l
a
WU2
b
WU4
x1
x1
y’
ys’
x
W
ys’’
Ws
WU1
as
tr
y’’
bs
WU3
56
Sprung Weight
Sprung Weight Distribution
• Taking the moments about the rear axle gives the longitudinal
location of the sprung mass CG.
(3542 * 48.7) (210 * 98)
bs
49.08in
3095
as 98 49.08 48.92in
Example C3 Corvette Upgrade:
WT = 3542lb
WS = 3095lb
WUF = 210lb
b = 48.7in
l = 98in
l
a
WU2
b
WU4
x1
x1
y’
ys’
x
W
ys’’
Ws
WU1
as
tr
y’’
bs
WU3
57
Sprung Weight
Sprung Weight Distribution
• If the font and rear unsprung weight are equal side to side, and the
front/rear tracks are the same, then the lateral location of the sprung
mass CG is found by taking the moments about the x1 axis as:
W
WUR
WUF
y' s T * y'
*t
*t
Ws
2 *Ws 2 *Ws
l
a
WU2
b
WU4
x1
x1
y’
ys’
x
W
ys’’
Ws
WU1
as
tr
y’’
bs
WU3
58
Sprung Weight
Sprung Weight Distribution
• If the font and rear unsprung weight are equal side to side, and the
front/rear tracks are the same, then the lateral location of the sprung
mass CG is found by taking the moments about the x1 axis as:
Example C3 Corvette Upgrade:
WT = 3542lb
WS = 3095lb
WUF = 210lb
WUR = 237lb
y’ = 29.33in
t = 58.66in
3542
237
210
y's
* 29.33
* 58.66
* 58.66
3095
2 * 3095
2 * 3095
y's 29.33in
l
a
WU2
b
WU4
x1
x1
y’
ys’
x
W
ys’’
Ws
WU1
as
tr
y’’
bs
WU3
59
Sprung Weight
Sprung Weight Distribution
• The sprung weight front and rear:
WSF WS *(as / ) Sprung Weight Front
WSR WS (bs / ) Sprung Weight Rear
l
a
WU2
b
WU4
x1
x1
y’
ys’
x
W
ys’’
Ws
WU1
as
tr
y’’
bs
WU3
60
Sprung Weight
Sprung Weight Distribution
• The sprung weight front and rear:
WSF 3095 * (48.92 / 98) 1544.4lb
Example C3 Corvette Upgrade:
WS = 3095lb
aS = 48.92in
bS = 49.08in
l = 98in
WSR 3095 * (49.08 / 98) 1550.6lb
l
a
WU2
b
WU4
x1
x1
y’
ys’
x
W
ys’’
Ws
WU1
as
tr
y’’
bs
WU3
61
Sprung and Unsprung Weight
Sprung Weight CG height
hS
(WT * h) (WUF * RLF ) (WUR * RLR )
WS
Where:
hS = Sprung weight CG height (in)
WT = Total vehicle weight (lb)
h = Vehicle CG height (in)
WUF = Unsprung weight front (lb)
RLF = Tire Loaded Radius front (in)
WUR = Unsprung weight rear (lb)
RLR = Tire Loaded Radius rear (in)
WS = Sprung weight (lb)
CGS
hs
CGUF
CGveh
CGUR
62
Sprung and Unsprung Weight
Sprung Weight CG height
Example C3 Corvette Upgrade:
hS = Sprung weight CG height (in)
WT = 3542lb
h = 17in
WUF = 210lb
RLF = 12.27in
WUR = 237lb
RLR = 12.83in
WS = 3095lb
(3542 * 17) (210 * 12.27) (237 * 12.83)
3095
hS 17.64in
hS
CGS
hs
CGUF
CGveh
CGUR
63
Roll Stiffness
Roll Stiffness
• Roll stiffness is the resistance of the springs within the suspension
against the body roll when a vehicle goes through a corner.
• Roll stiffness is developed by a roll resisting moment of the body
(sprung mass) about the roll axis.
• The roll stiffness in a vehicle is equal to the combined roll stiffness of
the front and rear suspension.
• Roll stiffness expressed in (torque) ft-lb/degree of roll.
• Example: If a vehicle had a roll stiffness of 600 ft-lb/deg of roll, it would
take a torque of 600 ft-lb about the roll axis to move the body 1 degree.
• Roll stiffness of the complete vehicle is the sum of the separate roll
stiffness rates of all vehicle suspensions.
• Tire deflection rates are included in the front and rear roll stiffness
rate values.
64
Roll Stiffness
Roll Stiffness
• Roll stiffness is the torque (T) (moment or roll couple) to rotate the
body (sprung weight) about the roll axis is shown in the following
equations.
Where:
T = Torque
KL = left vertical spring rate
KR = right vertical spring rate
t = distance between the springs
t
t t t
T = KL + KR
2 2 2 2
t2
T = (KL + KR)
4
Roll Axis
F
For equal spring rates, left and
right the above equation reduces
to the following:
2
T=
t
(K)
2
65
Roll Stiffness
Roll Stiffness
• Roll stiffness (KF) in radians for suspension with equal spring rate
either side (symmetric) is shown in the following equation.
2
Where:
KF = Roll Stiffness (Roll Rate)
T = Torque
K = vertical spring rate or wheel rate*
t = distance between the springs
2
t
t
T = (K) = (K)F
2
2
T
2
t
=
=
(K)
KF
2
Roll Axis
F
*Solid axles with leaf springs would
use vertical spring rates (K).
*Independent suspensions and antiroll bars would use the wheel rates (K).
66
Roll Stiffness
Roll Stiffness
• Roll stiffness (KF) expressed in metric units (N-m/deg).
2
t
t
K
K
KF
(2 * 180 ) 114.6
2
T
• Roll stiffness (KF) expressed in English units (ft-lb/deg).
KF
T
(2 *
t2
180
t2
K
K
1375
* 12)
Assuming original (t) values given in
inches and (K) values in lb/in.
67
Roll Moment
Sprung Weight Roll Moment lever arm
hRM
aS
hS Z F ( Z R Z F ) * (1 )
Where:
hRM = Sprung weight RM lever arm (in)
hS = Sprung weight CG height (in)
ZF = Roll Center height front (in)
ZR = Roll Center height rear (in)
aS/l = Sprung mass weight dist. front (%)
CGS
hRM
ZF
CGveh
Roll Axis
ZR
68
Roll Moment
Sprung Weight Roll Moment lever arm
hRM 17.64 .497 (.896 .497) * (1 .499)
hRM 17.19in
Example C3 Corvette Upgrade :
hRM = Sprung weight RM lever arm (in)
hS = 17.64in
ZF = 0.497in
ZR = 0.896in
aS/l = .499
CGS
hRM
ZF
CGveh
Roll Axis
ZR
69
Roll Moment
Roll Moment per lateral g acceleration
MF hRM *WS
Ay
12
Where:
M F = Roll Moment (ft-lb)
A y = Lateral acceleration (g)
hRM = Sprung weight RM lever arm (in)
WS = Sprung weight (lb)
F
Sprung wt
Cg
Roll Axis
hRM
Direction of Turn
70
Roll Moment
Roll Moment per lateral g acceleration
MF 17.19 * 3095
4433.6 ft lb / g
Ay
12
Example C3 Corvette Upgrade:
M F = Roll Moment (ft-lb)
A y = Lateral acceleration (g)
hRM = 17.19in
WS = 3095lb
F
Sprung wt
Cg
Roll Axis
hRM
Direction of Turn
71
Roll Stiffness
Total Roll Stiffness Rate per Roll Gradient
KF
MF / Ay
RG
Where:
K F = Total Roll Stiffness (ft-lb/deg)
M F = Roll Moment (ft-lb)
A y = Lateral acceleration (g)
RG = Roll Gradient = 1.5deg/g
F
Sprung wt
Cg
Roll Axis
hRM
Direction of Turn
72
Roll Stiffness
Total Roll Stiffness per Roll Gradient
KF
4433.6
2955.7 ft lb / deg
1.5
Example C3 Corvette Upgrade:
K F = Total Roll Stiffness (ft-lb/deg)
M F = 4433.6ft-lb
A y = 1g
RG = Roll Gradient = 1.5deg/g
F
Sprung wt
Cg
Roll Axis
hRM
Direction of Turn
73
Roll Stiffness
Front Roll Stiffness
• To calculate the available roll stiffness from the springs alone
for an independent suspension:
KFSF
Where:
K FSF = Front Roll Stiffness (ft-lb/deg)
K RF = Front Ride Rate (lb/in)
T F = Track front (in)
1375 = 2*(180/)*12
2
K RF * (TF )
1375
F
Sprung wt
Cg
Roll Axis
hRM
Direction of Turn
74
Roll Stiffness
Front Roll Stiffness
• To calculate the available roll stiffness from the springs alone
for an independent suspension:
200.77 * (58.66)2
KFSF
1375
Example C3 Corvette Upgrade:
K FSF = Front Roll Stiffness (ft-lb/deg)
K RF = 200.77lb/in
T F = 58.66in
1375 = 2*(180/)*12
KFSF 502.4 ft lb / deg
F
Sprung wt
Cg
Roll Axis
hRM
Direction of Turn
75
Roll Stiffness
Front Roll Stiffness
• To calculate the available roll stiffness from the springs alone
for an solid (live) axle suspension:
Where:
K FSF = Front Roll Stiffness (ft-lb/deg)
K WF = Front Wheel Rate (lb/in)
T S = Spring spacing (in)
K T = Tire Rate (lb/in)
T F = Track front (in)
1375 = 2*(180/)*12
(KWF * TS 2 ) * (KT * TF 2 )
KFSF
1375 * (KWF * TS 2 ) (KT * TF 2 )
F
Sprung wt
Cg
Roll Axis
hRM
Direction of Turn
76
Roll Stiffness
Rear Roll Stiffness
• To calculate the available roll stiffness from the springs alone
for an independent suspension:
KFSR
Where:
K FSR = Rear Roll Stiffness (ft-lb/deg)
K RR = Rear Ride Rate (lb/in)
T R = Track rear (in)
1375 = 2*(180/)*12
2
K RR * (TR )
1375
F
Sprung wt
Cg
Roll Axis
hRM
Direction of Turn
77
Roll Stiffness
Rear Roll Stiffness
• To calculate the available roll stiffness from the springs alone
for an independent suspension:
266.81 * (58.66)2
KFSR
1375
Example C3 Corvette Upgrade:
K FSR = Rear Roll Stiffness (ft-lb/deg)
K RR = 266.81lb/in
T R = 58.66in
1375 = 2*(180/)*12
KFSR 667.7 ft lb / deg
F
Sprung wt
Cg
Roll Axis
hRM
Direction of Turn
78
Roll Stiffness
Rear Roll Stiffness
• To calculate the available roll stiffness from the springs alone
for an solid (live) axle suspension:
Where:
K FSR = Rear Roll Stiffness (ft-lb/deg)
K WR = Rear Wheel Rate (lb/in)
T S = Spring spacing (in)
K T = Tire Rate (lb/in)
T R = Track rear (in)
1375 = 2*(180/)*12
(KWR * TS 2 ) * (KT * TR 2 )
KFSR
1375 * (KWR * TS 2 ) (KT * TR 2 )
F
Sprung wt
Cg
Roll Axis
hRM
Direction of Turn
79
Anti-Roll Bars
Anti-Roll Bars
• Total stiffness due to springs:
KFS KFSF KFSR
Where:
K FS = Total Spring Roll Stiffness (ft-lb/deg)
K FSF = Front Spring Roll Stiffness (ft-lb/deg)
K FSR = Rear Spring Roll Stiffness (ft-lb/deg)
• Anti-roll bars would then need to provide the difference to
equal the Total Roll Stiffness:
KFB KF KFS
Where:
K FB = Total ARB Roll Stiffness (ft-lb/deg)
K F = Total Roll Stiffness (ft-lb/deg)
K FS = Spring Roll Stiffness (ft-lb/deg)
80
Anti-Roll Bars
Anti-Roll Bars
• Total stiffness due to springs:
KFS 502.4 667.7
KFS 1170.1 ft lb / deg
Example C3 Corvette Upgrade:
K FS = Total Spring Roll Stiffness (ft-lb/deg)
K FSF = 502.4ft-lb/deg
K FSR = 667.7ft-lb/deg
• Anti-roll bars would then need to provide the difference to
equal the Total Roll Stiffness:
KFB 2955.7 1170.1
KFB 1785.6 ft lb / deg
Example C3 Corvette Upgrade:
K FB = Total ARB Roll Stiffness (ft-lb/deg)
K F = 2955.7ft-lb/deg
K FS = 1170.1ft-lb/deg
81
Anti-Roll Bars
Anti-Roll Bars
• To calculate the requirements of the front and rear anti-roll
bars, it is important to know the lateral load transfer
distribution per g of acceleration.
TLT WT * h
Ay
Tave
Where:
TLT = Total Load Transfer (lb)
A y = Lateral acceleration (g)
WT = Total vehicle weight (lb)
h = vehicle CG height (in)
Tave = Average track width [(TF+TR)/2] (in)
• To insure initial understeer of the vehicle, calculate the Front
Lateral Load Transfer to be 5% above the total front weight
distribution (WF + 5%).
TLT
* (WF 5%)
FLT
A
y
Where:
FLT = Front Load Transfer (lb)
TLT = Total Load Transfer (lb)
A y = Lateral acceleration (g)
WF = Front vehicle weight (lb)
82
Anti-Roll Bars
Anti-Roll Bars
• To calculate the requirements of the front and rear anti-roll
bars, it is important to know the lateral load transfer
distribution per g of acceleration.
TLT 3542 * 17
1026.49lb / g
Ay
58.66
Example C3 Corvette Upgrade:
TLT = Total Load Transfer (lb)
A y = Lateral acceleration (g)
WT = 3542lb
h = 17in
Tave = 58.66in
• To insure initial understeer of the vehicle, calculate the Front
Lateral Load Transfer to be 5% above the total front weight
distribution (WF% + 5%).
FLT 1026.49 * (.497 .05)
FLT 561.49lb
Example C3 Corvette Upgrade:
FLT = Front Load Transfer (lb)
TLT = 1026.49lb
A y = 1g
WF %= 49.7%
83
Anti-Roll Bars
Anti-Roll Bars
• To calculate the front body roll stiffness solve for K F :
FLT 12(K FF ) * F (WSF * ZF ) (WUF * RLF )
Ay
TF
TF
TF
F
Sprung wt
Cg
Roll Axis
Where:
FLT = Front Load Transfer (lb)
A y = Lateral acceleration (g)
K FF = Front Roll Stiffness (ft-lb/deg)
F = Roll gradient (deg)
WSF = Sprung weight front (lb)
ZF = Roll center height front (in)
WUF = Unsprung weight front (lb)
RLF = Tire static load radius front (in)
TF = Front track width (in)
Direction of Turn
84
Anti-Roll Bars
Anti-Roll Bars
• To calculate the front body roll stiffness solve for K F :
12(K FF ) * 1.5 (1544.4 * .497) (210 * 12.27)
58.66
58.66
58.66
561.49 0.307(K FF ) 13.09 43.93
561.49
K FF 504.47 / .307 1643.2 ft lb / deg
F
Sprung wt
Cg
Roll Axis
Direction of Turn
Example C3 Corvette Upgrade:
FLT = 561.49lb
A y = 1g
K FF = Front Roll Stiffness (ft-lb/deg)
F = 1.5deg
WSF = 1544.4lb
ZF = 0.497in
WUF = 210lb
RLF = 12.27in
TF = 58.66in
85
Anti-Roll Bars
Anti-Roll Bars
• To balance the body roll stiffness between the springs and the
ARB, the front anti-roll bar stiffness is calculated as:
KFBF KFF KFSF
Where:
K FBF = Front ARB Roll Stiffness Req’d (ft-lb/deg)
K FF = Front Roll Stiffness (ft-lb/deg)
K FSF = Front Spring Roll Stiffness (ft-lb/deg)
• To balance the body roll stiffness between the springs and the
ARB, the rear anti-roll bar stiffness is calculated as:
KFBR KF KFF KFSR
Where:
K FBR = Rear ARB Roll Stiffness Req’d (ft-lb/deg)
K F = Total Roll Stiffness (ft-lb/deg)
K FF = Front Roll Stiffness (ft-lb/deg)
K FSR = Rear Spring Roll Stiffness (ft-lb/deg)
86
Anti-Roll Bars
Anti-Roll Bars
• To balance the body roll stiffness between the springs and the
ARB, the front anti-roll bar stiffness is calculated as :
K FBF 1643.2 502.4
K FBF 1140.8 ft lb / deg
Example C3 Corvette Upgrade:
K FBF = Front ARB Roll Stiffness Req’d (ft-lb/deg)
K FF = 1643.2ft-lb/deg
K FSF = 502.4ft-lb/deg
• To balance the body roll stiffness between the springs and the
ARB, the rear anti-roll bar stiffness is calculated as :
K FBR 2955.7 1643.2 667.7
K FBR 644.8 ft lb / deg
Example C3 Corvette Upgrade:
K FBR = Rear ARB Roll Stiffness Req’d (ft-lb/deg)
K F = 2955.7ft-lb/deg
K FF = 1643.2ft-lb/deg
K FSR = 667.7ft-lb/deg
87
Anti-Roll Bars
Anti-Roll Bars
• The front anti-roll bar rate (lb/in) for body roll compensation
can be derived from the ARB stiffness as:
K FBF * 1375
K BF
(TF )2
Where:
K BF = Front ARB Body Roll Rate (lb/in)
K FBF = Front ARB Roll Stiffness (ft-lb/deg)
TF = Front track width(in)
1375 = (2*180/*12)
• The rear anti-roll bar rate (lb/in) for body roll compensation
can be derived from the ARB stiffness as :
K FBR * 1375
K BR
(TR )2
Where:
K BR = Rear ARB Body Roll Rate (lb/in)
K FBR = Rear ARB Roll Stiffness (ft-lb/deg)
TR = Rear track width(in)
1375 = (2*180/*12)
88
Anti-Roll Bars
Anti-Roll Bars
• The front anti-roll bar rate (lb/in) for body roll compensation
can be derived from the ARB stiffness as:
1140.8 * 1375
K BF
(58.66)2
K BF 455.8lb/in
Example C3 Corvette Upgrade:
K BF = Front ARB Body Roll Rate (lb/in)
K FBF = 1140.8ft-lb/deg
TF = 58.66in
1375 = (2*180/*12)
• The rear anti-roll bar rate (lb/in) for body roll compensation
can be derived from the ARB stiffness as :
644.8 * 1375
K BR
(58.66)2
K BR 257.6lb/in
Example C3 Corvette Upgrade:
K BR = Rear ARB Body Roll Rate (lb/in)
K FBR = 644.8ft-lb/deg
TR = 58.66in
1375 = (2*180/*12)
89
Anti-Roll Bars
Anti-Roll Bars
• The anti-roll bars must apply their force through their motion
ratios. Therefore:
• The front anti-roll bar rate is calculated as:
(K BF / 2)
K bf
(MR f )2
Where:
K bf = Front ARB Roll Rate (lb/in)
K BF = Front ARB Body Roll Rate (lb/in)
MRf = Front ARB motion ratio (in/in)
• The rear anti-roll bar rate is calculated as:
K br
(K BR / 2)
(MRr )2
Where:
K br = Rear ARB Roll Rate (lb/in)
K BR = Rear ARB Body Roll Rate (lb/in)
MRr = Rear ARB motion ratio (in/in)
90
Anti-Roll Bars
Anti-Roll Bars
• The anti-roll bars must apply their force through their motion
ratios. Therefore:
• The front anti-roll bar rate is calculated as:
(455.8 / 2)
K bf
(.643)2
Example C3 Corvette Upgrade:
K bf = Front ARB Roll Rate (lb/in)
K BF = 455.8lb/in
MRf = 0.643in/in
K bf 551.2lb/in
• The rear anti-roll bar rate is calculated as:
(257.6 / 2)
K bf
(.75)2
K bf 229lb/in
Example C3 Corvette Upgrade:
K br = Rear ARB Roll Rate (lb/in)
K BR = 257.6lb/in
MRr = 0.75in/in
91
Anti-Roll Bars
Anti-Roll Bars
• Anti-roll bars perform in torsion. The deflection rate at the free
end of a torsion spring is:
F
d 4G
32 L r
2
k
L
d
Where:
G = Modulus of Rigidity
= Deflection
r
r
92
Anti-Roll Bars
Anti-roll bars (independent suspension)
• The ARB stiffness (kind bar) [ft-lb/deg] for this type of anti-roll bar
(torsion bar) is calculated as:
k ind bar
(500,000)OD 4
(500,000)id 4
2
2
*
MR
*
2
*
(
t
)
r
2
3
2
3
(0.4422 A * B) 0.2264C (0.4422 A * B) 0.2264C
1375
When one wheel hits a bump the anti-roll bar twists as the wheel is raised, and since the other wheel does
not move, the bar twists over its whole length (B). In roll the bar is twisted from both ends so its effective
length is half the actual length which doubles the one wheel rate of the anti-roll bar.
93
Anti-Roll Bars
Anti-roll bars (solid axle)
• The ARB stiffness (kbar) [ft-lb/deg] for this type of anti-roll bar
(torsion bar) is calculated as:
2
D r2
2
*
2
*
(
t
)
1,178,000
r
2
r
L
A
1
k sol bar
1375
4
Where:
tr = track width (in)
(r2/r1) = Motion Ratio
When one wheel hits a bump the anti-roll bar twists as the wheel is raised, and since the other wheel does
not move, the bar twists over its whole length (L) . In roll the bar is twisted from both ends so its effective
length is half the actual length which doubles the one wheel rate of the anti-roll bar.
Pivot
r2
r1
94
Cornering Forces
Centripetal vs. Centrifugal Force
• When the trajectory of an object travels on a closed path about a
point -- either circular or elliptical -- it does so because there is a force
pulling the object in the direction of that point. That force is defined
as the CENTRIPETAL force.
• CENRTIFUGAL force is a force that operates in the opposite direction
as the CENTRIPETAL force. The centripetal force points inward toward the center of the turn (circle). The feeling of being "thrown
outward“ is due to the inertia of an object. Therefore, the inertial
reaction could be considered by some as centrifugal force.
95
Cornering Forces
Centripetal Force
WT v 2
Fc mac
*
ag
r
FN = mag = W T (vertical forces cancel)
ac = centripetal acceleration
3542lb
(51.33 ft / sec)2
Fc
*
966lb
2
32.2 ft / sec
300 ft
Example: C3 Corvette Upgrade
WT = 3542 lb
v = 35 mph = 51.33 ft/sec
r = 300 ft
ag = 32.2 ft/sec2
FN
r
Fc
WT = mag
96
Cornering Forces
Lateral Acceleration (g’s)
v2
ac as distance/sec2
r
v2 / r
ac
as g’s
ag
Example: C3 Corvette Upgrade
WT = 3542 lb
v = 35 mph = 51.33 ft/sec
r = 300 ft
ag = 32.2 ft/sec2
(51.33 ft/sec )2
ac
8.78 ft/sec 2
300 ft
(51.33 ft/sec )2 / 300 ft
ac
.273g' s
2
32.2 ft / sec
FN = mag = W T (vertical forces cancel)
ac = centripetal acceleration
ag = acceleration due to gravity (1g)
FN
r
Fc
WT = mag
97
Cornering Forces
Frictional Force
• If frictional force (Ff) is equal to centripetal force (Fc) the vehicle
would be at its limit of adhesion to the road.
WT v 2
Fc mac
*
ag
r
Ff mFN mmag
FN = mag = W T (vertical forces cancel)
ac = centripetal acceleration
m = Coefficient of Friction between tires and road
(dry pavement 0.7 – 0.8; wet pavement 0.3 – 0.4)
v2
velocity for a
mag v 2 mag r v mag r Max
given radius and m
r
v2
v2
radius for a
mag r
Min
r
mag given velocity and m
FN
r
Fc
Ff
WT = mag
98
Cornering Forces
Frictional Force
• If frictional force (Ff) is equal to centripetal force (Fc) the vehicle
would be at its limit of adhesion to the road.
3542 87.912
Fc
*
2833.6lb
32.2
300
Ff 0.8 * 3542 2833.6lb
FN = mag = W T (vertical forces cancel)
ac = centripetal acceleration
m = Coefficient of Friction between tires and road
(dry pavement 0.7 – 0.8; wet pavement 0.3 – 0.4)
velocity for a
v 0.8 * 32.2 * 300 87.91 ft/sec 59.9mph Max
given radius and m
51.332
r
102.3 ft
0.8 * 32.2
Min radius for a
given velocity and m
Example: C3 Corvette Upgrade
WT = 3542 lb
v = 35 mph = 51.33 ft/sec
r = 300 ft
ag = 32.2 ft/sec2
m = 0.8
FN
r
Fc
Ff
WT = mag
99
References:
1.
2.
3.
4.
5.
6.
Ziech, J., “Weight Distribution and Longitudinal Weight Transfer - Session 8,”
Mechanical and Aeronautical Engineering, Western Michigan University.
Hathaway, R. Ph.D, “Spring Rates, Wheel Rates, Motion Ratios and Roll
Stiffness,” Mechanical and Aeronautical Engineering, Western Michigan
University.
Gillespie, T. Ph.D, Fundamentals of Vehicle Dynamics, Society of Automotive
Engineers International, Warrendale, PA, February, 1992, (ISBN: 978-1-56091199-9).
Reimpell, J., Stoll, H., Betzler, J. Ph.D, The Automotive Chassis: Engineering
Principles, 2nd Ed., Butterworth-Heinemann, Woburn, MA, 2001, (ISBN 0 7506
5054 0).
Milliken, W., Milliken, D., Race Car Vehicle Dynamics, Society of Automotive
Engineers International, Warrendale, PA, February, 1994, (ISBN: 978-1-56091526-3).
Puhn, F., How to Make Your Car Handle, H.P. Books, Tucson, AZ, 1976 (ISBN 0912656-46-8).
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The End
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