The document discusses calculating vehicle roll properties and load transfer. It provides information needed to determine roll centers, roll axis, tire stiffness rates, sprung mass center of gravity location, and distribution of sprung weight between the front and rear of the vehicle. Parameters like roll center heights, track widths, spring rates, and vehicle weight distributions are required. Roll centers are defined as the intersection points of lines drawn from the center of the tire contact patch through the suspension. The total sprung weight is calculated by subtracting the unsprung weight from the total vehicle weight.
This is Part 3 of a 10 Part Series in Automotive Dynamics and Design, with an emphasis on Mass Properties. This series was intended to constitute the basis of a semester long course on the subject.
Longitudinal Vehicle Dynamics
-Maximum tractive effort of two-axle and track-semitrailer vehicles.
-The braking force of a two-axle vehicle.
-Acceleration time and distance.
-Relationship between engine torque and thrust force.
-Relationship between engine speed and vehicle speed
This is Part 3 of a 10 Part Series in Automotive Dynamics and Design, with an emphasis on Mass Properties. This series was intended to constitute the basis of a semester long course on the subject.
Longitudinal Vehicle Dynamics
-Maximum tractive effort of two-axle and track-semitrailer vehicles.
-The braking force of a two-axle vehicle.
-Acceleration time and distance.
-Relationship between engine torque and thrust force.
-Relationship between engine speed and vehicle speed
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.
This is Part 4 of a 10 Part Series in Automotive Dynamics and Design, with an emphasis on Mass Properties. This series was intended to constitute the basis of a semester long course on the subject.
To analyze the influence of tire properties on the dynamic behavior of vehicles, the engineer requires an accurate description of the tire-road contact phenomena
Forces are generated at the tire contact patch during various maneuvers of the car and transferred to the chassis through the suspension links. Calculating the forces on every link is important to design the suspension system as all the forces from wheel to the chassis are transferred by the suspension linkages. These forces have been calculated for all the links of a double wishbone suspension geometry. The load paths and FBD have been drawn and axial stress in the all the linkages
Roof Crush Analysis For occupant safety and ProtectionPratik Saxena
Optimization for Roof Crush Analysis under section FMVSS-216. Performed this test on the passenger’s side using Hypermesh and LS-Dyna placed the dummy (Hybrid III 50th percentile), seat, seat belt and side airbag on passenger’s side to perform the analysis. Performed optimization to reduce the chances of injury.
To provide good ride and handling performance –
–vertical compliance providing chassis isolation
–ensuring that the wheels follow the road profile
–very little tire load fluctuation
•To ensure that steering control is maintained during maneuvering –
–wheels to be maintained in the proper position wrt road surface
•To ensure that the vehicle responds favorably to control forces produced by the tires during
–longitudinal braking
–accelerating forces,
–lateral cornering forces and
–braking and accelerating torques
–this requires the suspension geometry to be designed to resist squat, dive and roll of the vehicle body
•To provide isolation from high frequency vibration from tire excitation
–requires appropriate isolation in the suspension joints
–Prevent transmission of ‘road noise’ to the vehicle body
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Wheel Dimensions, Wheel Balancing, and Wheel Alignment. Structure and Function of Tyres, Static and
Dynamic Properties of Pneumatic Tyres, Types of Tyres, Materials, Tyre Section & Designation, Factors
affecting Tyre Life, Tyre Rotation.
Bearings: Functions; classification of bearings; bearing materials; automotive bearings.
Design of half shaft and wheel hub assembly for racing carRavi Shekhar
The Half - Shaft and Wheel Hub of Formula One racing car was designed taking into consideration one of the popular model of Redbull racing car. The various dimension of shaft and hub were altered to attain maximum factor of safety.
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.
This is Part 4 of a 10 Part Series in Automotive Dynamics and Design, with an emphasis on Mass Properties. This series was intended to constitute the basis of a semester long course on the subject.
To analyze the influence of tire properties on the dynamic behavior of vehicles, the engineer requires an accurate description of the tire-road contact phenomena
Forces are generated at the tire contact patch during various maneuvers of the car and transferred to the chassis through the suspension links. Calculating the forces on every link is important to design the suspension system as all the forces from wheel to the chassis are transferred by the suspension linkages. These forces have been calculated for all the links of a double wishbone suspension geometry. The load paths and FBD have been drawn and axial stress in the all the linkages
Roof Crush Analysis For occupant safety and ProtectionPratik Saxena
Optimization for Roof Crush Analysis under section FMVSS-216. Performed this test on the passenger’s side using Hypermesh and LS-Dyna placed the dummy (Hybrid III 50th percentile), seat, seat belt and side airbag on passenger’s side to perform the analysis. Performed optimization to reduce the chances of injury.
To provide good ride and handling performance –
–vertical compliance providing chassis isolation
–ensuring that the wheels follow the road profile
–very little tire load fluctuation
•To ensure that steering control is maintained during maneuvering –
–wheels to be maintained in the proper position wrt road surface
•To ensure that the vehicle responds favorably to control forces produced by the tires during
–longitudinal braking
–accelerating forces,
–lateral cornering forces and
–braking and accelerating torques
–this requires the suspension geometry to be designed to resist squat, dive and roll of the vehicle body
•To provide isolation from high frequency vibration from tire excitation
–requires appropriate isolation in the suspension joints
–Prevent transmission of ‘road noise’ to the vehicle body
Wheels and Tyres: Types of Wheels, Construction, Structure and Function, Forces acting on wheels,
Wheel Dimensions, Wheel Balancing, and Wheel Alignment. Structure and Function of Tyres, Static and
Dynamic Properties of Pneumatic Tyres, Types of Tyres, Materials, Tyre Section & Designation, Factors
affecting Tyre Life, Tyre Rotation.
Bearings: Functions; classification of bearings; bearing materials; automotive bearings.
Design of half shaft and wheel hub assembly for racing carRavi Shekhar
The Half - Shaft and Wheel Hub of Formula One racing car was designed taking into consideration one of the popular model of Redbull racing car. The various dimension of shaft and hub were altered to attain maximum factor of safety.
DESIGN AND FABRICATION OF SINGLE REDUCTION GEARBOX WITH INBOARD BRAKINGabdul mohammad
An inboard braking system is an automobile technology where in the disc brakes are mounted on the chassis or to the gearbox of the vehicle, rather than directly on the wheel hubs.
The main advantages are a reduction in the unsprung weight of the wheel hubs, as this no longer includes the brake discs and calipers; also, braking torque applies directly to the chassis or the gear box , rather than being taken through the suspension arms.
Inboard brakes are fitted to a driven axle of the car, as they require a drive shaft to link the wheel to the brake. Most have thus been used for rear-wheel drive cars, although four-wheel drive and some front-wheel drives have also used them.
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"Trans Failsafe Prog" on your BMW X5 indicates potential transmission issues requiring immediate action. This safety feature activates in response to abnormalities like low fluid levels, leaks, faulty sensors, electrical or mechanical failures, and overheating.
Comprehensive program for Agricultural Finance, the Automotive Sector, and Empowerment . We will define the full scope and provide a detailed two-week plan for identifying strategic partners in each area within Limpopo, including target areas.:
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• Scope: Develop collaborations with automotive service providers to improve service quality and business operations.
• Target Areas: Polokwane, Lephalale, Mokopane, Phalaborwa, and Bela-Bela.
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• Scope: Provide business support support and training to women-owned businesses, promoting economic inclusion.
• Target Areas: Polokwane, Thohoyandou, Musina, Burgersfort, and Louis Trichardt.
We will also prioritize Industrial Economic Zone areas and their priorities.
Sign up on https://profilesmes.online/welcome/
To be eligible:
1. You must have a registered business and operate in Limpopo
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3. Sectors : Agriculture ( primary and secondary) and Automative
Women and Youth are encouraged to apply even if you don't fall in those sectors.
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Discover the reasons why your BMW’s Active Steering malfunction warning might come on. From electrical glitches to mechanical failures and software anomalies, addressing these promptly with professional inspection and maintenance ensures continued safety and performance on the road, maintaining the integrity of your driving experience.
Things to remember while upgrading the brakes of your carjennifermiller8137
Upgrading the brakes of your car? Keep these things in mind before doing so. Additionally, start using an OBD 2 GPS tracker so that you never miss a vehicle maintenance appointment. On top of this, a car GPS tracker will also let you master good driving habits that will let you increase the operational life of your car’s brakes.
Symptoms like intermittent starting and key recognition errors signal potential problems with your Mercedes’ EIS. Use diagnostic steps like error code checks and spare key tests. Professional diagnosis and solutions like EIS replacement ensure safe driving. Consult a qualified technician for accurate diagnosis and repair.
In this presentation, we have discussed a very important feature of BMW X5 cars… the Comfort Access. Things that can significantly limit its functionality. And things that you can try to restore the functionality of such a convenient feature of your vehicle.
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𝘼𝙣𝙩𝙞𝙦𝙪𝙚 𝙋𝙡𝙖𝙨𝙩𝙞𝙘 𝙏𝙧𝙖𝙙𝙚𝙧𝙨 𝙞𝙨 𝙫𝙚𝙧𝙮 𝙛𝙖𝙢𝙤𝙪𝙨 𝙛𝙤𝙧 𝙢𝙖𝙣𝙪𝙛𝙖𝙘𝙩𝙪𝙧𝙞𝙣𝙜 𝙩𝙝𝙚𝙞𝙧 𝙥𝙧𝙤𝙙𝙪𝙘𝙩𝙨. 𝙒𝙚 𝙝𝙖𝙫𝙚 𝙖𝙡𝙡 𝙩𝙝𝙚 𝙥𝙡𝙖𝙨𝙩𝙞𝙘 𝙜𝙧𝙖𝙣𝙪𝙡𝙚𝙨 𝙪𝙨𝙚𝙙 𝙞𝙣 𝙖𝙪𝙩𝙤𝙢𝙤𝙩𝙞𝙫𝙚 𝙖𝙣𝙙 𝙖𝙪𝙩𝙤 𝙥𝙖𝙧𝙩𝙨 𝙖𝙣𝙙 𝙖𝙡𝙡 𝙩𝙝𝙚 𝙛𝙖𝙢𝙤𝙪𝙨 𝙘𝙤𝙢𝙥𝙖𝙣𝙞𝙚𝙨 𝙗𝙪𝙮 𝙩𝙝𝙚 𝙜𝙧𝙖𝙣𝙪𝙡𝙚𝙨 𝙛𝙧𝙤𝙢 𝙪𝙨.
Over the 10 years, we have gained a strong foothold in the market due to our range's high quality, competitive prices, and time-lined delivery schedules.
1. BND TechSource
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Vehicle Load Transfer
Part III of III
Lateral Load Transfer
Wm Harbin
Technical Director
BND TechSource
(revised 04JUL21)
2. BND TechSource
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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)
2
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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.
3
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Roll Centers (continued):
▪ 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.
4
Zrc
Roll
Center
Zrc = 16 in
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Roll Centers (continued):
▪ 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.
5
Zr
Roll
Center
Control Arms
Zr = 0.896 in
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Roll Centers (continued):
▪ 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.
6
Zf
Roll
Center
Control Arms
Ball Joints
Ball Joints
To IC To IC
Zf = 0.497 in
7. BND TechSource
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Roll Axis
▪ A theoretical line connecting the two roll centers is called the roll axis.
7
Rear Roll Center
Front Roll Center
Roll Axis
Vehicle Center of Gravity
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Tire Stiffness Rate
▪ Tires in a suspension system not only control the grip to the road surface,
they also act as additional springs (and dampers) within the system.
▪ There are many ways to calculate tire stiffness 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.
8
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Tire Stiffness Rate (continued):
▪ These examples are for the static rate of the tires shown.
▪ The tire stiffness rate will change due to changes in air pressure (Δ1 bar can affect
rate by 40%) and slightly (less than 1%) when the vehicle speed changes.
9
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
in
lb
in
in
lb
kt /
3
.
1411
03
.
13
)
2
/
95
.
28
(
3
.
2039
=
−
=
in
lb
in
in
lb
kt /
2
.
1560
87
.
11
)
2
/
69
.
25
(
2
.
1521
=
−
=
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
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
in
lb
in
in
lb
kt /
4
.
1761
36
.
12
)
2
/
66
.
26
(
6
.
1708
=
−
=
10. BND TechSource
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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).
10
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
in
RL 77
.
13
)
3
.
1411
/
992
(
)
2
/
95
.
28
( =
−
=
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
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
in
RLF 27
.
12
)
2
.
1560
/
893
(
)
2
/
69
.
25
( =
−
=
in
RLR 83
.
12
)
4
.
1761
/
882
(
)
2
/
66
.
26
( =
−
=
11. BND TechSource
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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%.
11
Right Front
Unsprung
Cg
Left Rear
Unsprung
Cg
Body
Sprung
Cg
Left Front
Unsprung
Cg
Right Rear
Unsprung
Cg
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Sprung and Unsprung Weight (continued):
▪ The sprung weight of the vehicle is simply the total weight minus the unsprung
weight.
12
Weight
Sprung
W
W
W
W
W
W U
U
U
U
s =
−
−
−
−
= 4
3
2
1
tr
WU1
WU2
WU3
WU4
as bs
a b
l
x
y’
y’’
ys’
ys’’ W
W
s
x1
x1
13. BND TechSource
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Sprung and Unsprung Weight (continued):
▪ The sprung weight of the vehicle is simply the total weight minus the unsprung
weight.
13
tr
WU1
WU2
WU3
WU4
as bs
a b
l
x
y’
y’’
ys’
ys’’ W
W
s
x1
x1
lb
W
W
S
s
3095
5
.
118
5
.
118
105
105
3542
=
−
−
−
−
=
Example C3 Corvette Upgrade:
WS = Sprung weight (lb)
WT = 3542lb
WU1 = 105lb
WU2 = 105lb
WU3 = 118.5lb
WU4 = 118.5lb
14. BND TechSource
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Sprung Weight Distribution
▪ The unsprung weight front and rear:
14
Rear
Weight
Unsprung
W
W
W
Front
Weight
Unsprung
W
W
W
U
U
UR
U
U
UF
=
+
=
=
+
=
4
3
2
1
tr
WU1
WU2
WU3
WU4
as bs
a b
l
x
y’
y’’
ys’
ys’’ W
Ws
x1
x1
15. BND TechSource
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Sprung Weight Distribution (continued):
▪ The unsprung weight front and rear:
15
tr
WU1
WU2
WU3
WU4
as bs
a b
l
x
y’
y’’
ys’
ys’’ W
Ws
x1
x1
lb
W
lb
W
UR
UF
237
5
.
118
5
.
118
210
105
105
=
+
=
=
+
=
Example C3 Corvette Upgrade:
WUF = Sprung weight front (lb)
WUR = Sprung weight rear (lb)
WU1 = 105lb
WU2 = 105lb
WU3 = 118.5lb
WU4 = 118.5lb
16. BND TechSource
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Sprung Weight Distribution (continued):
▪ Taking the moments about the rear axle gives the longitudinal location of the sprung
mass CG (WS).
16
tr
WU1
WU2
WU3
WU4
as bs
a b
l
x
y’
y’’
ys’
ys’’
W
Ws
x1
x1
s
UF
T
s
W
W
b
W
b
)
*
(
)
*
(
−
= and s
s b
a −
=
17. BND TechSource
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Sprung Weight Distribution (continued):
▪ Taking the moments about the rear axle gives the longitudinal location of the sprung
mass CG (WS).
17
tr
WU1
WU2
WU3
WU4
as bs
a b
l
x
y’
y’’
ys’
ys’’
W
Ws
x1
x1
in
bs 08
.
49
3095
)
98
*
210
(
)
7
.
48
*
3542
(
=
−
=
in
as 92
.
48
08
.
49
98 =
−
=
Example C3 Corvette Upgrade:
WT = 3542lb
WS = 3095lb
WUF = 210lb
b = 48.7in
l = 98in
18. BND TechSource
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Sprung Weight Distribution (continued):
▪ 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 (WS) is found by
taking the moments about the x1 axis as:
18
tr
WU1
WU2
WU3
WU4
as bs
a b
l
x
y’
y’’
ys’
ys’’
W
Ws
x1
x1
−
−
= t
W
W
t
W
W
y
W
W
y
s
UF
s
UR
s
T
s *
*
2
*
*
2
'
*
'
19. BND TechSource
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Sprung Weight Distribution (continued):
▪ 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 (WS) is found by
taking the moments about the x1 axis as:
19
tr
WU1
WU2
WU3
WU4
as bs
a b
l
x
y’
y’’
ys’
ys’’
W
Ws
x1
x1
in
y
y
s
s
33
.
29
'
66
.
58
*
3095
*
2
210
66
.
58
*
3095
*
2
237
33
.
29
*
3095
3542
'
=
−
−
=
Example C3 Corvette Upgrade:
WT = 3542lb
WS = 3095lb
WUF = 210lb
WUR = 237lb
y’ = 29.33in
t = 58.66in
20. BND TechSource
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Sprung Weight Distribution (continued):
▪ The sprung weight front and rear:
20
tr
WU1
WU2
WU3
WU4
as bs
a b
l
x
y’
y’’
ys’
ys’’
W
Ws
x1
x1
Rear
Weight
Sprung
b
W
W
Front
Weight
Sprung
a
W
W
s
S
SR
s
S
SF
=
+
=
=
=
)
/
(
)
/
(
*
21. BND TechSource
BND TechSource https://bndtechsource.wixsite.com/home
Sprung Weight Distribution (continued):
▪ The sprung weight front and rear:
21
tr
WU1
WU2
WU3
WU4
as bs
a b
l
x
y’
y’’
ys’
ys’’
W
Ws
x1
x1
lb
W
lb
W
SR
SF
6
.
1550
)
98
/
08
.
49
(
*
3095
4
.
1544
)
98
/
92
.
48
(
*
3095
=
=
=
=
Example C3 Corvette Upgrade:
WS = 3095lb
aS = 48.92in
bS = 49.08in
l = 98in
22. BND TechSource
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Sprung Weight CG height
22
CGveh
CGS
CGUF CGUR
hs
S
R
UR
F
UF
T
S
W
RL
W
RL
W
h
W
h
)
*
(
)
*
(
)
*
( −
−
=
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)
24. BND TechSource
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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.
24
25. BND TechSource
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Roll Stiffness (continued):
▪ 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.
25
2
t
K
2
t
+
2
t
K
2
t
=
T R
L
)
K
+
K
(
4
t
=
T R
L
2
(K)
2
t
=
T
2
For equal spring rates, left and right
the above equation reduces to the
following:
Where:
T = Torque
KL = left vertical spring rate
KR = right vertical spring rate
t = distance between the springs
Roll Axis F
26. BND TechSource
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Roll Stiffness (continued):
▪ Roll stiffness (Kᶲ) in radians for suspension with equal spring rate either side
(symmetric) is shown in the following equation.
26
Roll Axis F
F
(K)
2
t
=
(K)
2
t
=
T
2
2
Where:
KF = Roll Stiffness (Roll Rate)
T = Torque
K = vertical spring rate or wheel rate*
t = distance between the springs
(K)
2
t
=
T
=
K
2
F
*Solid axles with leaf springs would use
vertical spring rates (K).
*Independent suspensions and anti-roll
bars would use the wheel rates (K).
27. BND TechSource
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Roll Stiffness (continued):
▪ Roll stiffness (Kᶲ) expressed in metric units (N-m/deg).
▪ Roll stiffness (Kᶲ) expressed in imperial units (ft-lb/deg).
27
Assuming original (t) values given in mm and (K) values in N/mm.
114600 = 2*(180/π)*1000
K
t
K
t
T
K
2
1375
)
12
*
180
*
2
(
2
=
=
=
F
Assuming original (t) values given in inches and (K) values in lb/in.
1375 = 2*(180/π)*12
28. BND TechSource
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Roll Moment
▪ Sprung Weight Roll Moment lever arm
28
−
−
+
−
= )
1
(
*
)
(
S
F
R
F
S
RM
a
Z
Z
Z
h
h
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 (%)
CGveh
CGS
ZF
ZR
hRM
Roll Axis
29. BND TechSource
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Roll Moment (continued):
▪ Sprung Weight Roll Moment lever arm
29
CGveh
CGS
ZF
ZR
hRM
Roll Axis
in
h
h
RM
RM
19
.
17
)
499
.
1
(
*
)
497
.
896
(.
497
.
64
.
17
=
−
−
+
−
=
Example C3 Corvette Upgrade :
hRM = Sprung weight RM lever arm (in)
hS = 17.64in
ZF = 0.497in
ZR = 0.896in
aS/l = .499
30. BND TechSource
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Roll Moment (continued):
▪ Roll Moment per lateral g acceleration
30
Where:
M F = Roll Moment (ft-lb)
Ay = Lateral acceleration (g)
hRM = Sprung weight RM lever arm (in)
WS = Sprung weight (lb)
12
*
1
*
)
(g
S
RM
y
W
h
A
M
=
F
Roll Axis
Sprung wt
Cg
Direction of Turn
F
hRM
31. BND TechSource
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Roll Moment (continued):
▪ Roll Moment per lateral g acceleration
31
Roll Axis
Sprung wt
Cg
Direction of Turn
F
hRM
Example C3 Corvette Upgrade:
M F = Roll Moment (ft-lb)
Ay = Lateral acceleration (g)
hRM = 17.19in
WS = 3095lb
g
lb
ft
A
M
y
/
6
.
4433
12
3095
*
19
.
17
−
=
=
F
32. BND TechSource
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Roll Stiffness
▪ Total Roll Stiffness Rate per (desired) Roll Gradient
32
Where:
K F = Total Roll Stiffness (ft-lb/deg)
M F = Roll Moment (ft-lb)
Ay = Lateral acceleration (g)
RG = Roll Gradient = 1.5deg/g
Note: RG target of 1.5 deg/g was requested by
the customer.
RG
A
M
K
y
/
F
F =
Roll Axis
Sprung wt
Cg
Direction of Turn
F
hRM
33. BND TechSource
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Roll Stiffness (continued):
▪ Total Roll Stiffness Rate per (desired) Roll Gradient
33
Example C3 Corvette Upgrade:
K F = Total Roll Stiffness (ft-lb/deg)
M F = 4433.6ft-lb
Ay = 1g
RG = Roll Gradient = 1.5deg/g
Note: RG target of 1.5 deg/g was
requested by the customer.
deg
/
7
.
2955
5
.
1
6
.
4433
lb
ft
K −
=
=
F
Roll Axis
Sprung wt
Cg
Direction of Turn
F
hRM
34. BND TechSource
BND TechSource https://bndtechsource.wixsite.com/home
Roll Stiffness (continued):
▪ Front Roll Stiffness
▪ To calculate the available roll stiffness from the springs alone for an
independent suspension:
34
Roll Axis
Sprung wt
Cg
Direction of Turn
F
hRM
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
1375
)
(
* 2
F
RF
SF
T
K
K =
F
35. BND TechSource
BND TechSource https://bndtechsource.wixsite.com/home
Roll Stiffness (continued):
▪ Front Roll Stiffness
▪ To calculate the available roll stiffness from the springs alone for an
independent suspension:
35
Roll Axis
Sprung wt
Cg
Direction of Turn
F
hRM
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
deg
/
4
.
502
1375
)
66
.
58
(
*
77
.
200 2
lb
ft
K
K
SF
SF
−
=
=
F
F
36. BND TechSource
BND TechSource https://bndtechsource.wixsite.com/home
Roll Stiffness (continued):
▪ Front Roll Stiffness
▪ To calculate the available roll stiffness from the springs alone for an solid (live)
axle suspension:
36
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
)
*
(
)
*
(
*
1375
)
*
(
*
)
*
(
2
2
2
2
F
T
S
WF
F
T
S
WF
SF
T
K
T
K
T
K
T
K
K
+
=
F
Sprung wt
Cg
Roll
Axis
Direction of Turn
F
hRM
37. BND TechSource
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Roll Stiffness (continued):
▪ Rear Roll Stiffness
▪ To calculate the available roll stiffness from the springs alone for an
independent suspension:
37
Roll Axis
Sprung wt
Cg
Direction of Turn
F
hRM
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
1375
)
(
* 2
R
RR
SR
T
K
K =
F
38. BND TechSource
BND TechSource https://bndtechsource.wixsite.com/home
Roll Stiffness (continued):
▪ Rear Roll Stiffness
▪ To calculate the available roll stiffness from the springs alone for an
independent suspension:
38
Roll Axis
Sprung wt
Cg
Direction of Turn
F
hRM
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
deg
/
7
.
667
1375
)
66
.
58
(
*
81
.
266 2
lb
ft
K
K
SR
SR
−
=
=
F
F
39. BND TechSource
BND TechSource https://bndtechsource.wixsite.com/home
Roll Stiffness (continued):
▪ Rear Roll Stiffness
▪ To calculate the available roll stiffness from the springs alone for an solid (live)
axle suspension:
39
Sprung wt
Cg
Roll
Axis
Direction of Turn
F
hRM
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
)
*
(
)
*
(
*
1375
)
*
(
*
)
*
(
2
2
2
2
R
T
S
WR
R
T
S
WR
SR
T
K
T
K
T
K
T
K
K
+
=
F
40. BND TechSource
BND TechSource https://bndtechsource.wixsite.com/home
Roll Gradient
▪ Roll Gradient w/o ARB
▪ To calculate the roll gradient from the springs alone without ARB:
40
Roll Axis
Sprung wt
Cg
Direction of Turn
F
hRM
Where:
φnoARB= Roll Gradient w/o ARB (deg/g)
Ws = Sprung Weight (lb)
hRM= Roll moment arm length (in)
K φSF = Front Roll Stiffness (ft-lb/deg)
K φSR = Rear Roll Stiffness (ft-lb/deg)
41. BND TechSource
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Roll Gradient (continued):
▪ Roll Gradient w/o ARB
▪ To calculate the roll gradient from the springs alone without ARB:
41
Roll Axis
Sprung wt
Cg
Direction of Turn
F
hRM
Where:
φnoARB= Roll Gradient w/o ARB (deg/g)
Ws = 3095 (lb)
hRM= 17.19(in)
K φSF = 502.4 (ft-lb/deg)
K φSR = 667.7 (ft-lb/deg)
42. BND TechSource
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Anti-Roll Bars
▪ Total stiffness due to springs:
▪ Anti-roll bars would then need to provide the difference to equal the Total Roll
Stiffness:
42
SR
SF
S K
K
K F
F
F +
=
S
B K
K
K F
F
F −
=
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)
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)
43. BND TechSource
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Anti-Roll Bars (continued):
▪ Total stiffness due to springs:
▪ Anti-roll bars would then need to provide the difference to equal the Total Roll
Stiffness:
43
deg
/
1
.
1170
7
.
667
4
.
502
lb
ft
K
K
S
S
−
=
+
=
F
F
deg
/
6
.
1785
1
.
1170
7
.
2955
lb
ft
K
K
B
B
−
=
−
=
F
F
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
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
44. BND TechSource
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Anti-Roll Bars (continued):
▪ 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.
▪ To insure initial understeer of the vehicle, calculate the Front Lateral Load
Transfer to be 5% above the total front weight distribution (WF + 5%).
44
ave
T
y T
h
W
A
TLT *
=
Where:
TLT = Total Load Transfer (lb)
Ay = Lateral acceleration (g)
WT = Total vehicle weight (lb)
h = vehicle CG height (in)
Tave = Average track width [(TF+TR)/2] (in)
%)
5
(
* +
= F
y
W
A
TLT
FLT
Where:
FLT = Front Load Transfer (lb)
TLT = Total Load Transfer (lb)
Ay = Lateral acceleration (g)
WF = Front vehicle weight (lb)
45. BND TechSource
BND TechSource https://bndtechsource.wixsite.com/home
Anti-Roll Bars (continued):
▪ 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.
▪ To insure initial understeer of the vehicle, calculate the Front Lateral Load
Transfer to be 5% above the total front weight distribution (WF + 5%).
45
g
lb
A
TLT
y
/
49
.
1026
66
.
58
17
*
3542
=
=
Example C3 Corvette Upgrade:
TLT = Total Load Transfer (lb)
Ay = Lateral acceleration (g)
WT = 3542lb
h = 17in
Tave = 58.66in
lb
FLT
FLT
49
.
561
)
05
.
497
(.
*
49
.
1026
=
+
= Example C3 Corvette Upgrade:
FLT = Front Load Transfer (lb)
TLT = 1026.49lb
Ay = 1g
WF %= 49.7%
46. BND TechSource
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Anti-Roll Bars (continued):
▪ To calculate the front body roll stiffness solve for KᶲF :
46
F
F
UF
F
F
SF
F
F
y T
RL
W
T
Z
W
T
K
A
FLT )
*
(
)
*
(
*
)
(
12
+
+
F
= F
Where:
FLT = Front Load Transfer (lb)
Ay = Lateral acceleration (g)
K FF = Front Roll Stiffness (ft-lb/deg)
F = (desired) Roll gradient (deg/g)
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)
F
F
UF
F
F
SF
F
F
y T
RL
W
T
Z
W
K
T
A
FLT )
*
(
)
*
(
*
*
12
+
+
F
=
F
F
+
−
=
F
F
F
F
UF
F
F
SF
y
F
T
T
RL
W
T
Z
W
A
FLT
K
*
12
)
*
(
)
*
(
47. BND TechSource
BND TechSource https://bndtechsource.wixsite.com/home
Anti-Roll Bars (continued):
▪ To calculate the front body roll stiffness solve for KᶲF :
47
deg
/
2
.
1643
307
.
/
47
.
504
93
.
43
09
.
13
)
(
307
.
0
49
.
561
66
.
58
)
27
.
12
*
210
(
66
.
58
)
497
.
*
4
.
1544
(
66
.
58
5
.
1
*
)
(
12
49
.
561
lb
ft
K
K
K
F
F
F
−
=
=
+
+
=
+
+
=
F
F
F
Example C3 Corvette Upgrade:
FLT = 561.49lb
Ay = 1g
K FF = Front Roll Stiffness (ft-lb/deg)
F = 1.5deg/g
WSF = 1544.4lb
ZF = 0.497in
WUF = 210lb
RLF = 12.27in
TF = 58.66in
48. BND TechSource
BND TechSource https://bndtechsource.wixsite.com/home
Anti-Roll Bars (continued):
▪ To balance the body roll stiffness between the springs and the ARB, the front
anti-roll bar stiffness is calculated as:
▪ To balance the body roll stiffness between the springs and the ARB, the rear
anti-roll bar stiffness is calculated as:
48
SF
F
BF K
K
K F
F
F −
= 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)
SR
F
BR K
K
K
K F
F
F
F −
−
= 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)
49. BND TechSource
BND TechSource https://bndtechsource.wixsite.com/home
Anti-Roll Bars (continued):
▪ To balance the body roll stiffness between the springs and the ARB, the front
anti-roll bar stiffness is calculated as:
▪ To balance the body roll stiffness between the springs and the ARB, the rear
anti-roll bar stiffness is calculated as:
49
deg
/
8
.
1140
4
.
502
2
.
1643
lb
ft
K
K
BF
BF
−
=
−
=
F
F
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
deg
/
8
.
644
7
.
667
2
.
1643
7
.
2955
lb
ft
K
K
BR
BR
−
=
−
−
=
F
F
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
50. BND TechSource
BND TechSource https://bndtechsource.wixsite.com/home
Anti-Roll Bars (continued):
▪ The front anti-roll bar suspension rate (lb/in) for body roll compensation can be
derived from the ARB stiffness as:
▪ The rear anti-roll bar suspension rate (lb/in) for body roll compensation can be
derived from the ARB stiffness as :
50
2
)
(
1375
*
F
BF
BF
T
K
K F
=
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)
2
)
(
1375
*
R
BR
BR
T
K
K F
=
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)
51. BND TechSource
BND TechSource https://bndtechsource.wixsite.com/home
Anti-Roll Bars (continued):
▪ The front anti-roll bar suspension rate (lb/in) for body roll compensation can be
derived from the ARB stiffness as:
▪ The rear anti-roll bar suspension rate (lb/in) for body roll compensation can be
derived from the ARB stiffness as :
51
lb/in
K
K
BF
BF
8
.
455
)
66
.
58
(
1375
*
8
.
1140
2
=
=
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)
lb/in
K
K
BR
BR
6
.
257
)
66
.
58
(
1375
*
8
.
644
2
=
=
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)
52. BND TechSource
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Anti-Roll Bars (continued):
▪ The anti-roll bars must apply their force through their motion ratios. Therefore:
▪ The front anti-roll bar rate is calculated as:
▪ The rear anti-roll bar rate is calculated as:
52
2
)
(
)
2
/
(
f
BF
bf
MR
K
K =
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)
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)
2
)
(
)
2
/
(
r
BR
br
MR
K
K =
53. BND TechSource
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Anti-Roll Bars (continued):
▪ The anti-roll bars must apply their force through their motion ratios. Therefore:
▪ The front anti-roll bar rate is calculated as:
▪ The rear anti-roll bar rate is calculated as:
53
lb/in
K
K
bf
bf
2
.
551
)
643
(.
)
2
/
8
.
455
(
2
=
=
Example C3 Corvette Upgrade:
K bf = Front ARB Roll Rate (lb/in)
K BF = 455.8lb/in
MRf = 0.643in/in
Example C3 Corvette Upgrade:
K br = Rear ARB Roll Rate (lb/in)
K BR = 257.6lb/in
MRr = 0.75in/in
lb/in
K
K
bf
bf
229
)
75
(.
)
2
/
6
.
257
(
2
=
=
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Check Roll Gradient
▪ Roll Gradient w/ ARB
▪ To calculate the roll gradient from the full suspension including the ARB’s:
54
Where:
Φw/ARB= Roll Gradient w/ ARB (deg/g)
Ws = Sprung Weight (lb)
hRM= Roll moment arm length (in)
K φSF = Front Roll Stiffness (ft-lb/deg)
K φSR = Rear Roll Stiffness (ft-lb/deg)
K φBF = Front Roll Stiffness (ft-lb/deg)
K φBR = Rear Roll Stiffness (ft-lb/deg)
55. BND TechSource
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Check Roll Gradient
▪ Roll Gradient w/ ARB
▪ To calculate the roll gradient from the full suspension including the ARB’s:
55
Where:
Φw/ARB= Roll Gradient w/ ARB (deg/g)
Ws = 3095 (lb)
hRM= 17.19(in)
K φSF = 502.4 (ft-lb/deg)
K φSR = 667.7 (ft-lb/deg)
K φBF = 1140.8 (ft-lb/deg)
K φBR = 644.8 (ft-lb/deg)
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Anti-Roll Bars
▪ Anti-roll bars perform in torsion. The deflection rate at the free end of
a torsion spring is:
56
Where:
G = Modulus of Rigidity
= Deflection
L
r
d
k
r
L
G
d
F
=
= 2
4
32
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Anti-roll bars (independent suspension)
▪ The ARB stiffness (kᶲind bar) [ft-lb/deg] for this type of anti-roll bar
(torsion bar) is calculated as:
57
1375
)
(
*
2
*
*
2264
.
0
)
*
4422
.
0
(
)
000
,
500
(
2264
.
0
)
*
4422
.
0
(
)
000
,
500
( 2
2
3
2
4
3
2
4
r
t
MR
C
B
A
id
C
B
A
OD
k bar
ind
+
−
+
=
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.
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Anti-roll bars (solid axle)
▪ The ARB stiffness (kᶲsol bar) [ft-lb/deg] for this type of anti-roll bar
(torsion bar) is calculated as:
58
1375
)
(
*
2
*
000
,
178
,
1 2
2
1
2
2
4
r
t
r
r
A
L
D
k bar
sol
=
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
r1
r2
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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.
59
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Cornering Forces (continued):
▪ Centripetal Force
60
r
v
a
W
ma
F
g
T
c
c
2
*
=
=
FN = mag = WT (vertical forces cancel)
ac = centripetal acceleration
Example: C3 Corvette Upgrade
WT = 3542 lb
v = 35 mph = 51.33 ft/sec
r = 300 ft
ag = 32.2 ft/sec2
lb
ft
ft
ft
lb
Fc 966
300
sec)
/
33
.
51
(
*
sec
/
2
.
32
3542 2
2
=
=
WT = mag
61. BND TechSource
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Cornering Forces (continued):
▪ Lateral Acceleration (g’s)
61
WT = mag
=
=
=
=
a
r
v
a
r
v
a
g
c
c
/
2
2
FN = mag = WT (vertical forces cancel)
ac = centripetal acceleration
ag = acceleration due to gravity (1g)
Example: C3 Corvette Upgrade
WT = 3542 lb
v = 35 mph = 51.33 ft/sec
r = 300 ft
ag = 32.2 ft/sec2
as distance/sec2
as g’s s
g
ft
ft
ft/sec
a
ft/sec
ft
ft/sec
a
c
c
'
273
.
sec
/
2
.
32
300
/
)
33
.
51
(
78
.
8
300
)
33
.
51
(
2
2
2
2
=
=
=
=
62. BND TechSource
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Cornering Forces (continued):
▪ Friction Coefficient
▪ Friction coefficient between the tires and the road surface depends on many
factors:
62
Vehicle speed
Tire loading (Peak & Sliding)
Road type/condition
Camber thrust
Slip angles
Tire construction and material compounds
63. BND TechSource
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Cornering Forces (continued):
▪ Frictional Force
▪ If frictional force (Ff) is equal to centripetal force (Fc) the vehicle would be at its
limit of adhesion to the road.
63
WT = mag
FN = mag = WT (vertical forces cancel)
ac = centripetal acceleration
µ = Coefficient of Friction between tires and road
(dry pavement 0.7 – 0.9; wet pavement 0.2 – 0.4
depending on speed & load)
g
N
f ma
F
F
=
=
=
=
=
=
=
=
=
g
g
g
g
g
a
v
r
r
v
a
r
a
v
r
a
v
r
v
a
2
2
2
2
Max velocity for a
given radius and µ
Min radius for a
given velocity and
r
v
a
W
ma
F
g
T
c
c
2
*
=
=
64. BND TechSource
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Cornering Forces (continued):
▪ Frictional Force
▪ If frictional force (Ff) is equal to centripetal force (Fc) the vehicle would be at its
limit of adhesion to the road.
64
WT = mag
FN = mag = WT (vertical forces cancel)
ac = centripetal acceleration
µ = Coefficient of Friction between tires and road
(dry pavement 0.7 – 0.9; wet pavement 0.2 – 0.4
depending on speed & load)
Max velocity for a
given radius and µ
Min radius for a
given velocity and
Example: C3 Corvette Upgrade
WT = 3542 lb
v = 55 mph = 80.67 ft/sec
r = 300 ft
ag = 32.2 ft/sec2
µ = 0.86
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Cornering Forces (continued):
▪ Cornering Force
▪ In the previous calculation we assumed all four tires at the same coefficient of
friction. However we know from previous roll calculations that each tire would
be loaded differently. Hence, the µ for each tire must be considered.
65
For more information please go to
https://bndtechsource.wixsite.com/home/c3-project-
docs and see: Susp_Project_Tasks_Calculator.xlsx
66. BND TechSource
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References:
1. Ziech, J., “Weight Distribution and Longitudinal Weight Transfer - Session 8,”
Mechanical and Aeronautical Engineering, Western Michigan University.
2. Hathaway, R. Ph.D, “Spring Rates, Wheel Rates, Motion Ratios and Roll Stiffness,” Mechanical
and Aeronautical Engineering, Western Michigan University.
3. Gillespie, T. Ph.D, Fundamentals of Vehicle Dynamics, Society of Automotive Engineers
International, Warrendale, PA, February, 1992, (ISBN: 978-1-56091-199-9).
4. 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).
5. Milliken, W., Milliken, D., Race Car Vehicle Dynamics, Society of Automotive Engineers
International, Warrendale, PA, February, 1994, (ISBN: 978-1-56091-526-3).
6. Puhn, F., How to Make Your Car Handle, H.P. Books, Tucson, AZ, 1976 (ISBN 0-912656-46-8).
66
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- END -
Vehicle Load Transfer
Part III of III
Lateral Load Transfer