Vehicle load transfer part III 2021

BND TechSource
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Vehicle Load Transfer
Part III of III
Lateral Load Transfer
Wm Harbin
Technical Director
BND TechSource
(revised 04JUL21)

BND TechSource
▪ 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

BND TechSource
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

BND TechSource
▪ 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

BND TechSource
▪ 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

BND TechSource
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

BND TechSource
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

BND TechSource
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)
Example 3:
Tire = P 275/40 R 18 99 W
Static (max) load radius = 314mm (12.36in)
in
lb
in
in
lb
kt /
4
.
1761
36
.
12
)
2
/
66
.
26
(
6
.
1708
=
−
=

BND TechSource
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
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
KT = 1560.2lb/in
Example 3:
Tire = P 275/40 R 18 99 W
KT = 1761.4lb/in
in
RLF 27
.
12
)
2
.
1560
/
893
(
)
2
/
69
.
25
( =
−
=
in
RLR 83
.
12
)
4
.
1761
/
882
(
)
2
/
66
.
26
( =
−
=

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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

BND TechSource
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

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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

BND TechSource
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
=
+
=
=
+
=
WUF = Sprung weight front (lb)
WUR = Sprung weight rear (lb)
WU1 = 105lb
WU2 = 105lb
WU3 = 118.5lb
WU4 = 118.5lb

BND TechSource
▪ 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 −
= 

BND TechSource
▪ 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 =
−
=
WT = 3542lb
WS = 3095lb
WUF = 210lb
b = 48.7in
l = 98in

BND TechSource
▪ 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
'
*
'

BND TechSource
▪ 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
'
=






−






−






=
WT = 3542lb
WS = 3095lb
WUF = 210lb
WUR = 237lb
y’ = 29.33in
t = 58.66in

BND TechSource
▪ 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
=
+
=
=
=
)
/
(
)
/
(
*



BND TechSource
▪ 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
=
=
=
=
WS = 3095lb
aS = 48.92in
bS = 49.08in
l = 98in

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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)

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Sprung Weight CG height (continued):
23
CGveh
CGS
CGUF CGUR
hs
in
h
h
S
S
64
.
17
3095
)
83
.
12
*
237
(
)
27
.
12
*
210
(
)
17
*
3542
(
=
−
−
=
WT = 3542lb
h = 17in
WUF = 210lb
RLF = 12.27in
WUR = 237lb
RLR = 12.83in
WS = 3095lb

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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

BND TechSource
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

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▪ 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).

BND TechSource
▪ 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

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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)
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

BND TechSource
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 :
hS = 17.64in
ZF = 0.497in
ZR = 0.896in
aS/l = .499

BND TechSource
30
Where:
M F = Roll Moment (ft-lb)
Ay = Lateral acceleration (g)
12
*
1
*
)
(g
S
RM
y
W
h
A
M
=
F
Roll Axis
Sprung wt
Cg
Direction of Turn
F
hRM

BND TechSource
31
Roll Axis
Sprung wt
Cg
Direction of Turn
F
hRM
hRM = 17.19in
WS = 3095lb
g
lb
ft
A
M
y
/
6
.
4433
12
3095
*
19
.
17
−
=
=
F

BND TechSource
Roll Stiffness
▪ Total Roll Stiffness Rate per (desired) Roll Gradient
32
Where:
K F = Total Roll Stiffness (ft-lb/deg)
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

BND TechSource
▪ Total Roll Stiffness Rate per (desired) Roll Gradient
33
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

BND TechSource
▪ 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

BND TechSource
35
Roll Axis
Sprung wt
Cg
Direction of Turn
F
hRM
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

BND TechSource
▪ To calculate the available roll stiffness from the springs alone for an solid (live)
axle suspension:
36
Where:
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

BND TechSource
▪ Rear Roll Stiffness
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

BND TechSource
38
Roll Axis
Sprung wt
Cg
Direction of Turn
F
hRM
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

BND TechSource
▪ 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 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

BND TechSource
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)

BND TechSource
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)

BND TechSource
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 FS = Spring Roll Stiffness (ft-lb/deg)

BND TechSource
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
K FS = Total Spring Roll Stiffness (ft-lb/deg)
K FSF = 502.4ft-lb/deg
K FSR = 667.7ft-lb/deg
K FB = Total ARB Roll Stiffness (ft-lb/deg)
K F = 2955.7ft-lb/deg
K FS = 1170.1ft-lb/deg

BND TechSource
▪ 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)
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)
WF = Front vehicle weight (lb)

BND TechSource
▪ 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
=
=
WT = 3542lb
h = 17in
Tave = 58.66in
lb
FLT
FLT
49
.
561
)
05
.
497
(.
*
49
.
1026
=
+
= Example C3 Corvette Upgrade:
TLT = 1026.49lb
Ay = 1g
WF %= 49.7%

BND TechSource
▪ 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:
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
)
*
(
)
*
(

BND TechSource
▪ 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
FLT = 561.49lb
Ay = 1g
F = 1.5deg/g
WSF = 1544.4lb
ZF = 0.497in
WUF = 210lb
RLF = 12.27in
TF = 58.66in

BND TechSource
▪ 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
48
SF
F
BF K
K
K F
F
F −
= Where:
K FBF = Front ARB Roll Stiffness Req’d (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 FSR = Rear Spring Roll Stiffness (ft-lb/deg)

BND TechSource
▪ To balance the body roll stiffness between the springs and the ARB, the front
▪ To balance the body roll stiffness between the springs and the ARB, the rear
49
deg
/
8
.
1140
4
.
502
2
.
1643
lb
ft
K
K
BF
BF
−
=
−
=
F
F
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
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

BND TechSource
▪ 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)

BND TechSource
▪ 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
=
=
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
=
=
K FBR = 644.8ft-lb/deg
TR = 58.66in
1375 = (2*180/*12)

BND TechSource
▪ 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)
MRf = Front ARB motion ratio (in/in)
Where:
K br = Rear ARB Roll Rate (lb/in)
MRr = Rear ARB motion ratio (in/in)
2
)
(
)
2
/
(
r
BR
br
MR
K
K =

BND TechSource
▪ 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
=
=
K bf = Front ARB Roll Rate (lb/in)
K BF = 455.8lb/in
MRf = 0.643in/in
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
=
=

BND TechSource
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)

BND TechSource
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)

BND TechSource
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



BND TechSource
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.

BND TechSource
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

BND TechSource
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

BND TechSource
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

BND TechSource
▪ Lateral Acceleration (g’s)
61
WT = mag
=
=
=
=
a
r
v
a
r
v
a
g
c
c
/
2
2
ag = acceleration due to gravity (1g)
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
=
=
=
=

BND TechSource
▪ 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

BND TechSource
▪ 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
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
*
=
=

BND TechSource
▪ 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
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 
WT = 3542 lb
v = 55 mph = 80.67 ft/sec
r = 300 ft
ag = 32.2 ft/sec2
µ = 0.86

BND TechSource
▪ 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

BND TechSource
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

BND TechSource
- END -
Vehicle Load Transfer
Part III of III

Vehicle load transfer part III 2021

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