Vehicle load transfer part II 2021

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

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Center of Gravity
2

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Center of Gravity Location
▪ Locating the center of gravity in the X-Y (horizontal) plane is
performed by placing the vehicle on scales and identifying the
corresponding loads.
▪ X, Y, positions are noted
▪ % Front, % Rear, % Left, % Right, % Diagonal (RF,LR)
▪ Locating the center of gravity height (hcg) can be achieved by
raising one end of the vehicle and identifying the load change on
the un-raised end which is a result of the height change on the
raised end.
▪ To avoid suspension “flex” adversely affecting the measurements,
use a solid link between the front unsprung to sprung components.
3

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Center of Gravity Location (continued):
WR
l
R
l
F
L
Cg
WF
SCALE SCALE
Center of Gravity
Horizontal Plane Location
4

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100
% 







 +
=
tot
LF
RF
front
W
W
W
W
100
% 







 +
=
tot
RR
LR
rear
W
W
W
W
100
% 







 +
=
tot
LR
LF
left
W
W
W
W 100
% 







 +
=
tot
RR
RF
right
W
W
W
W
100
% 







 +
=
tot
LR
RF
diag
W
W
W
W
5

BND TechSource
100
3542
880
880
%
7
.
49 





 +
=
= front
W
100
3542
891
891
%
3
.
50 





 +
=
= rear
W
100
3542
891
880
%
50 





 +
=
= left
W
100
3542
891
880
%
50 





 +
=
= right
W
100
3542
891
880
%
50 





 +
=
= diag
W
Example: C3 Corvette Upgrade
Weight total = 3542 lb
W RF = 880 lb
W LF = 880 lb
W RR = 891 lb
W LR = 891 lb
6

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L
W
W
L
W
W
tot
r
tot
f
f 
=









−
= 1

L
W
W
L
W
W
tot
f
tot
r
r 
=









−
= 1

7

BND TechSource
98
3542
1782
98
3542
1760
1
in
3
.
49 
=







−
=
= f

98
3542
1760
98
3542
1782
1
in
7
.
48 
=







−
=
= r

W tot = 3542 lb
W F = 1760 lb
W R = 1782 lb
L = 98.0 in
8

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Center of Gravity Height Location
 −
=
= L
W
WL
M
Horizontal f
R
R 0
:
( )







cos
cos
cos
sin
0
cos
cos
sin
0
:
L
W
L
W
L
W
h
W
or
L
W
W
L
W
h
W
M
Raised
f
f
R
cg
f
f
R
cg
R

−
−
+
=

+
−
+
=
=

To avoid suspension “flex”
adversely affecting the
measurements, use a solid link
between the front unsprung to
sprung components.
9

BND TechSource
Center of Gravity Height Location (continued):
The center of gravity height
above the spindle
centerline is:
h
W
L
W
W
L
W
above
h
tot
f
tot
f
cg


=

=
*
*
sin
*
cos
*
* 2





cos
sin
tan =

=
L
h
hCGtotal on level ground is:
(where rt = tire radius) t
tot
f
cg r
h
W
L
W
total
h +


=
*
* 2
10

BND TechSource
Center of Gravity Height Location (continued):
The center of gravity height
above the spindle
centerline is:
hCGtotal on level ground is:
(where rt = tire radius)
12
*
3542
9604
*
62
.
22
98
.
6
sin
*
3542
98
.
6
cos
*
98
*
62
.
22
11
.
5 =
=
98
.
6
cos
98
.
6
sin
98
12
98
.
6
tan =
=
89
.
11
12
*
3542
9604
*
62
.
22
in
17 +
=
Example: C3 Corvette
W tot = 3542 lb
W F = 1760 lb
W R = 1782 lb
L = 98.0 in
rt = 11.89 in
h = 12 in
WF= 22.62 lb
11

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12

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▪ Longitudinal (vehicle fore-aft direction) load transfer occurs due to
either positive (acceleration) or negative (braking) acceleration.
▪ Load transfer is associated with each of these accelerations. This
is due to the acceleration forces acting between the tire contact
patches at the road surface and the vehicle center of gravity height
which is above the road surface.
13

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Longitudinal Load Transfer (continued):
The vehicle load distribution on level ground is shown in the
following equations:
A B
WF
WR
l
R
l
F L
Cg
WT






=
L
W
=
W
W
=
L
W
Axle
Front R
F
R
F








=
L
W
=
W
W
=
L
W
Axle
Rear F
R
F
R


14

BND TechSource
The vehicle load distribution on level ground is shown in the
following equations:
A B
WF
WR
l
R
l
F L
Cg
WT
lb
=
W
W
=
L
W
Axle
Front F
R
F 1760
98
7
.
48
3542 =






= 
lb
=
W
W
=
L
W
Axle
Rear R
F
R 1782
98
3
.
49
3542 =






= 
W T = 3542 lb
W F = 1760 lb
W R = 1782 lb
L = 98.0 in
lF = 49.3 in
lR = 48.7 in
15

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The load transfer can be most easily determined on level ground, at
a speed low enough such that aerodynamic resistance force would
be negligible (zero). The equations are then solved for the static load
at each axle and any load transferred due to acceleration or
deceleration.
A B
WF +/- W WR +/- W
l
R
l
F L
acceleration
force
Cg
accel Fa
WT
hcg
L
h
a
a
W
L
h
a
m
=
L
h
F
=
W cg
g
x
T
cg
cg
a
T *
*
=

16

BND TechSource
The load transfer can be most easily determined on level ground, at
a speed low enough such that aerodynamic resistance force would
be negligible (zero). The equations are then solved for the static load
at each axle and any load transferred due to acceleration or
deceleration.
A B
WF +/- W WR +/- W
l
R
l
F L
acceleration
force
Cg
accel Fa
WT
hcg
lb
=
L
h
F
=
W cg
a
T 2
.
286
98
17
*
)
2
.
32
/
15
(
*
3542 =

W T = 3542 lb
L = 98.0 in
hcg = 17 in
ax = 15 ft/sec2
ag = 32.2 ft/sec2
17

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In forward acceleration the load on the front axle can be found by
solving the moment about point B of the figure.
A B
WF +/- W WR +/- W
l
R
l
F L
acceleration
force
Cg
accel Fa
WT
hcg






−

−

L
h
a
a
L
W
=
W
W
=
0
=
M(B) cg
g
x
R
T
F *

18

BND TechSource
In forward acceleration the load on the front axle can be found by
solving the moment about point B of the figure.
A B
WF +/- W WR +/- W
l
R
l
F L
acceleration
force
Cg
accel Fa
WT
hcg
W T = 3542 lb
W F = 1760 lb
W = 286.2 lb
l R = 48.7 in
L = 98.0 in
hcg = 17 in
ax = 15 ft/sec2
ag = 32.2 ft/sec2
lb
=
W
W
=
0
=
M(B) F 8
.
1473
98
17
*
2
.
32
15
98
7
.
48
3542 =






−

−

Check: 1473.8 lbs + 286.2 lbs = 1760 lb
19

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In forward acceleration the load on the rear axle can be found by
solving the moment about point A of the figure.
A B
WF +/- W WR +/- W
l
R
l
F L
acceleration
force
Cg
accel Fa
WT
hcg






+

+

L
h
a
a
L
W
=
W
W
=
0
=
M(A) cg
g
x
F
T
R *

20

BND TechSource
In forward acceleration the load on the rear axle can be found by
solving the moment about point A of the figure.
A B
WF +/- W WR +/- W
l
R
l
F L
acceleration
force
Cg
accel Fa
WT
hcg
W T = 3542 lb
W R = 1782 lb
W = 286.2 lb
l F = 49.3 in
L = 98.0 in
hcg = 17 in
ax = 15 ft/sec2
ag = 32.2 ft/sec2
lbs
=
W
W
=
0
=
M(A) R 2
.
2068
98
17
*
2
.
32
15
98
3
.
49
3542 =






+

+

Check: 2068.2 lbs - 286.2 lbs = 1782 lbs
21

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Longitudinal Acceleration Pitch
▪ The vehicle pitch angle θ, if no suspension forces oppose it (no
Anti-dive, anti-lift, or Anti-squat), is a function of the load transfer
(ΔW) and the wheel rate (KW).
▪ If the front wheel rate is KWf and the rear is KWr then the load transfer
(ΔW) would be divided between the left and right wheels. Therefore
vertical displacements (Z) of the sprung (body) mass at the wheel
locations are as shown.
ZR ZF
 
Cg
Cg
JOUNCE
JOUNCE
REBOUND
REBOUND
(Jounce and Rebound are also known as Bump and Droop)
22

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Longitudinal Acceleration Pitch (continued):
ZR ZF
 
Cg
Cg
JOUNCE
JOUNCE
REBOUND
REBOUND
(no Anti-dive, anti-lift, or Anti-squat)
K
W
=
Z
K
W
=
Z
Wf
F
Wr
R
2
/
2
/ 




180
L
K
W
+
K
W
=
L
Z
+
Z
=
L
K
W
+
K
W
= Wr
Wf
deg
F
R
Wr
Wf
rad *
2
/
2
/
2
/
2
/ 



23

BND TechSource
ZR ZF
 
Cg
Cg
JOUNCE
JOUNCE
REBOUND
REBOUND
(no Anti-dive, anti-lift, or Anti-squat)
in
=
Z
in
=
Z F
R 62
.
26
.
230
2
/
2
.
286
455
.
35
.
314
2
/
2
.
286
=
=
deg
63
.
*
98
26
.
230
2
/
2
.
286
35
.
314
2
/
2
.
286
01097
.
98
62
.
455
.
98
26
.
230
2
/
2
.
286
35
.
314
2
/
2
.
286
=
=



180
+
=
+
=
+
=
deg
rad
rad
KSf = 400 lb/in
MRf = 0.759 in/in
KWf = 230.26 lb/in
KSr = 600 lb/in
MRr = 0.724 in/in
KWr = 314.35 lb/in
W = 286.2 lb
L = 98.0 in
KW = MR2 * KS
24

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Suspension Geometry
25

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Instant Centers
▪ Instant Center (IC)
▪ Simply put, an instant center is a point in space (either real or
extrapolated) around which the suspension's links rotate.
▪ “Instant" means at that particular position of the linkage.
▪ "Center" refers to an extrapolated point that is the effective pivot
point of the linkage at that instant.
▪ The IC is used in both side view swing arm (SVSA) and a front view
swing arm (FVSA) geometry for suspension travel.
Front View Geometry
Side View Geometry
26

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Swing Arms
▪ Swing Arms
▪ There are many different types of vehicle suspension designs. All of
which have instant centers (reaction points) developed by running
lines through their pivots to an intersection point.
▪ A swing arm by definition has an minimum of two pivot points
attaching a suspension component to the vehicle chassis or
underbody. To simplify the concept, imagine a line running from the
IC directly to the suspension component. This line is referred to as
the swing arm.
FVSA
Swing Arm
SVSA
Swing Arm
27

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Swing Arms (continued):
▪ Swing Arms
▪ The side view swing arm controls force and motion factors
predominantly related to longitudinal accelerations, while the front
view swing arm controls force and motion factors due to lateral
accelerations.
FVSA
Swing Arm
SVSA
Swing Arm
28

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Side View Swing Arm
▪ Shown is a schematic of a solid rear drive axle with the linkages replaced
with a swing arm. In a solid drive axle the axle and differential move
together and are suspended from the chassis using springs and/or trailing
arms.
Cg
e
d
L
Fza
Fxa
Fx
Fz
IC
h
b
29

BND TechSource
Side View Swing Arm (continued):
▪ Shown is a schematic of an independent rear suspension (IRS) with the
linkages replaced with a swing arm. In an IRS the differential is mounted
to the chassis as the half shafts move independently and are suspended
from the chassis using springs, control arms and trailing arms.
Cg
e
d
L
Fza
Fxa
Fx
Fz
IC
h
b
r
30

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Anti- Geometry
31

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Anti- Geometry
▪ Anti-squat
Anti-squat in rear suspensions reduces the jounce (upward) travel
during forward acceleration on rear wheel drive cars only.
▪ Anti-dive
Anti-dive geometry in front suspensions reduces the jounce
(upward) deflection under forward braking.
▪ Anti-lift
Anti-lift in rear suspensions reduces rebound (downward) travel
during forward braking.
32

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Anti-squat Geometry
33

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Anti-squat Geometry
▪ Anti-squat
During forward (longitudinal) acceleration the vehicle load transfer tends
to compress the rear springs (suspension jounce) and allow the front
springs to extend (suspension rebound). Anti-squat characteristics can
be designed into the rear suspension geometry.
Anti-squat geometry produces a side view swing arm (SVSA) that
predicts suspension component behavior.

Cg
JOUNCE
REBOUND
34

BND TechSource
Anti-squat Geometry (continued):
▪ Anti-squat
Geometry that produces an instant center (IC) through which
acceleration forces (Fza and Fxa) can act to reduce or eliminate drive
wheel spring deflection during acceleration.
Cg
e
d
L
Fza
Fxa
Fx
Fz
IC
h
b
r
WT
100% Anti-squat
Line
35

BND TechSource
▪ Instant center locations are projected onto the longitudinal axis of
the vehicle. This provides the location where the forces transmitted
during acceleration effectively act.
▪ The resultant horizontal and vertical components of the tractive
force transmitted through these instant centers determine the load
percentage transfer that acts through the suspension linkages, with
the remaining load acting through the suspension springs.
36

BND TechSource
▪ The percentage of Anti-squat is now determined relative to the
100% (h/L) angle.
▪ If a suspension has 100% Anti-squat, all the longitudinal load
transfer is carried by the suspension linkages and not by the
springs (h/L line would be parallel to the swing arm line).
L
h
=
d
r
e
=
−
b
tan
Cg
e
d
L
Fza
Fxa
Fx
Fz
IC
h
b
r
WT
100% Anti-squat
Line
b
37

BND TechSource
▪ The magnitude of the vertical (Fza) components determines, in part,
the ability for the driver to accelerate without spinning the tires.
▪ The magnitude of the vertical (Fza) components also dictates the
load % transfer that is transmitted through the springs and
dampers to the road surface and the load % that transfers directly
through the suspension linkages.
38

BND TechSource
▪ The load transfer during acceleration is as shown:
▪ The tractive effort (Fx) at the drive wheels is calculated as:
L
h
a
a
W
=
L
h
a
m
=
F x
g
T
za
x
g
T
x
xa a
a
W
=
F
=
F
(1)
(2)
39

BND TechSource
▪ By examining the free body of the rear suspension and summing
moments about the contact patch, the equation below is derived.
▪ Since Fxa is the Tractive Force.
d
r
e
F
=
F
0
=
r
e
F
-
d
F
0
=
M
xa
za
xa
za
R
−
−

(3)
d
r
e
a
a
W
=
F x
g
T
za
−
(4)
Cg
e
d
L
Fza
Fxa
Fx
Fz
IC
h
b
r
WT
100% Anti-squat
Line
40

BND TechSource
▪ Equating equations (1) and (4) results in an equation which
indicates the relationships for 100% Anti-squat.
▪ The angle the instant center (IC) must lie on for 100% Anti-squat is:
d
r
e
=
L
h
d
r
e
a
a
W
=
L
h
a
a
W
=
F x
g
T
x
g
T
za
−

−
L
h
=
d
r
e
=
−
b
tan
41

BND TechSource
▪ If the tan β < h/L then squat will occur.
Cg
e
d
L
Fza
Fxa
Fx
Fz
IC
h
b
r
WT
100% Anti-squat
Line
L
h
=
d
r
e
=
−
b
tan
42

BND TechSource
▪ Anti-squat rear solid axle
Shown is a SVSA of a solid rear drive axle. The torque reaction is taken
by the suspension components.
Cg
e
d
L
Fza
Fxa
Fx
Fz
IC
h
b
L
h
d
e
=
squat
Anti =
%
100
100
*
/
/
/
tan
%
L
h
d
e
L
h
=
squat
Anti =
b
43

BND TechSource
▪ Anti-squat rear solid axle
Shown is a SVSA of a solid rear drive axle. The torque reaction is taken
by the suspension components.
Cg
e
d
L
Fza
Fxa
Fx
Fz
IC
h
b
%
125
100
*
100
/
20
45
/
25
.
11
/
tan
% =
=
L
h
=
squat
Anti
b
L
h
d
e
=
squat
Anti =
%
100
Example: Solid Axle
e = 11.25 in
d = 45 in
L = 100 in
h = 20 in
44

BND TechSource
▪ Anti-squat independent rear suspension
Shown is a SVSA of an independent rear suspension (IRS). The torque
reaction is taken by the chassis.
L
h
d
r
e
=
squat
Anti =
−
%
100
100
*
/
/
/
tan
%
L
h
d
r
e
L
h
=
squat
Anti
−
=
b
Cg
e
d
L
Fza
Fxa
Fx
Fz
IC
h
b
r
WT
100% Anti-squat
Line
45

BND TechSource
▪ Anti-squat independent rear suspension
Shown is a SVSA of an independent rear suspension (IRS). The torque
reaction is taken by the chassis.
Cg
e
d
L
Fza
Fxa
Fx
Fz
IC
h
b
r
WT
100% Anti-squat
Line
L
h
d
r
e
=
squat
Anti =
−
%
100
Example: C3 Upgrade
e = 15.62 in
d = 32.84 in
L = 98 in
h = 17 in
r = 11.89 in
%
5
.
65
100
*
98
/
17
84
.
32
/
89
.
11
62
.
15
/
tan
% =
−
=
L
h
=
squat
Anti
b
46

BND TechSource
▪ Anti-squat effects on longitudinal pitch angle
100
*
%
*
=
on
compensati
pitch
Wf
Wr
K
K
L
h
L
h
d
r
e






+
−
































+
−
−
L
h
K
d
r
e
K
L
h
K Wf
Wr
Wr
x
g
T
a
a
W
L
=
(rad)
angle
Pitch *
2
*
1
*
2
*
1
*
2
*
1
*
*
1

47

BND TechSource
▪ Anti-squat effects on longitudinal pitch angle
e – r = 3.73 in WT = 3542 lb
d = 32.84 in KWr = 314.35 lb/in
L = 98 in KWf = 230.26 lb/in
h = 17 in ax = 180 in/sec2
r = 11.89 in ag = 386.4 in/sec2















+






−



98
17
*
)
2
*
26
.
230
(
1
84
.
32
73
.
3
*
)
2
*
35
.
314
(
1
98
17
*
)
2
*
35
.
314
(
1
180
*
4
.
386
3542
*
98
1
=
(rad)
angle
Pitch 
%
28
100
*
4103
.
1136
.
% =
=
on
compensati
pitch
squat
Anti
deg
rad
=
(deg)
angle
Pitch %
5
.
65
@
45
.
0
180
*
0079
.
0 =


48
Check: (1-.28)*.63deg = .45deg

BND TechSource
ZR ZF
 
Cg
Cg
JOUNCE
JOUNCE
REBOUND
REBOUND
(no Anti-dive, anti-lift, or Anti-
squat)
in
=
Z
in
=
Z F
R 62
.
26
.
230
2
/
2
.
286
455
.
35
.
314
2
/
2
.
286
=
=
KSf = 400 lb/in
MRf = 0.759 in/in
KWf = 230.26 lb/in
KSr = 600 lb/in
MRr = 0.724 in/in
KWr = 314.35 lb/in
W = 286.2 lb
L = 98.0 in
KW = MR2 * KS
49
Previous
Slide No. 24
deg
63
.
*
98
26
.
230
2
/
2
.
286
35
.
314
2
/
2
.
286
01097
.
98
62
.
455
.
98
26
.
230
2
/
2
.
286
35
.
314
2
/
2
.
286
=
=



180
+
=
+
=
+
=
deg
rad
rad
Check: if e-r/d = 0,
then Anti-squat = 0
0.63˚ @ 0% Anti-squat

BND TechSource
Anti-dive Geometry
50

BND TechSource
Anti-dive Geometry
▪ Anti-dive
During braking (longitudinal deceleration) the vehicle load transfer tends
to compress the front springs (suspension jounce) and allow the rear
springs to extend (suspension rebound). Anti-dive is usually designed
into both front and rear suspensions (Anti-dive at the front and Anti-lift in
the rear).
Anti-dive geometry produces a side view swing arm (SVSA) that predicts
suspension component behavior.
51

Cg
JOUNCE
REBOUND

BND TechSource
Anti-dive Geometry (continued):
▪ Anti-dive
The total longitudinal load transfer under steady acceleration or braking
is a function of the wheelbase (L), CG height (h), and braking force
(WT)*(ax /ag).
52
Cg
L
h
WT
WT (ax/ag) = braking force
+  load
-  load
L
h
a
a
W
=
L
h
a
m
=
load x
g
T


BND TechSource
▪ Anti-dive
The total longitudinal load transfer under steady acceleration or braking
is a function of the wheelbase (L), CG height (h), and braking force
(WT)*(ax /ag).
53
lb
=
L
h
a
m
=
load 2
.
286
98
17
*
180
*
4
.
386
3542
=

WT = 3542 lb
L = 98 in
h = 17 in
ax = 180 in/sec2 (.46 g)
ag = 386.4 in/sec2

BND TechSource
Anti-dive
Geometry
(continued):
54
WT = 3542 lb
L = 98 in
h = 17 in
0
500
1000
1500
2000
2500
3000
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
Axle
Load
(lbf)
Deceleration (g's)
Load Transfer vs. Deceleration
Front Load (lbf)
Rear Load (lbf)
L
h
a
a
W
=
L
h
a
m
=
load x
g
T


BND TechSource
▪ Brake Bias (brake force distribution)
▪ The following factors will affect the load on an axle for any given
moment in time:
▪ Weight distribution of the vehicle (static).
▪ CG height – the higher it is, the more load transference during braking.
▪ Wheelbase – the shorter it is, the more load transference during braking.
▪ The following factors will affect how much brake torque is developed
at each corner of the vehicle, and how much of that torque is
transferred to the tire contact patch and reacted against the ground:
▪ Rotor effective diameter
▪ Caliper piston diameter
▪ Lining friction coefficients
▪ Tire traction coefficient properties
55

Cg
JOUNCE
REBOUND

BND TechSource
▪ Brake Bias (brake force distribution)
▪ Braking force at the tire contact patch vs. the total load on that tire will
determine the braking bias.
▪ Changing the CG height, wheelbase, or deceleration level will dictate
a different force distribution, or bias, requirement for a braking system.
▪ Conversely, changing the effectiveness of the front brake components
without changing the rear brake effectiveness can also cause our
brake bias to change.
56

Cg
JOUNCE
REBOUND

BND TechSource
Anti-dive
Geometry
(continued):
▪ Brake Bias
(brake force
distribution)
57

Cg
JOUNCE
REBOUND

BND TechSource
▪ Anti-dive (front) and Anti-lift (rear) suspension
▪ Shown is an SVSA with lines (100% Anti-dive/lift) representing the load transfer
during braking. If the IC’s are below these lines, the percentage of anti will be
below 100%. If the IC’s are above these lines, the percentage of anti will be
above 100%.
58
100
*
)
(%
*
)
/
(
*
tan
% braking
front
h
L
=
dive
Anti f
b
Cg
L
ICr
h
br
WT
ax
bf
ICf
%FB x L
100% Anti-dive
Line
100% Anti-lift
Line
1 - %FB x
L
100
*
)
%
1
(
*
)
/
(
*
tan
% braking
front
h
L
=
lift
Anti r −
b

BND TechSource
▪ Shown is an SVSA with lines (100% Anti-dive/lift) representing the load transfer
during braking. If the IC’s are below these lines, the percentage of anti will be
below 100%. If the IC’s are above these lines, the percentage of anti will be
above 100%.
59
Cg
L
ICr
h
br
WT
ax
bf
ICf
%FB x L
100% Anti-dive
Line
100% Anti-lift
Line
1 - %FB x
L
tan bf = .1124
tan br = .4894
L = 98 in
h = 17 in
% front braking @ .46g = .71
%
46
100
*
)
71
(.
*
)
765
.
5
(
*
1124
.
% =
=
dive
Anti
%
82
100
*
)
29
(.
*
)
765
.
5
(
*
4894
.
% =
=
lift
Anti

BND TechSource
▪ Since Anti-dive and Anti-lift are a resultant of braking force, and the braking
force changes due to brake bias, the % Anti changes as the rate of
deceleration changes.
60
0%
20%
40%
60%
80%
100%
120%
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
%
Anti-dive
&
Anti-lift
%
of
Total
Braking
Force
Deceleration (g's)
Anti-dive & Anti-lift vs. Brake Bias
% Anti-dive
% Front Braking
% Anti-lift
% Rear Braking

BND TechSource
Design factors in Anti-dive and Anti-Squat
▪ Since load transfer is a function of deceleration rate, and the brake forces are
shared, anti-dive geometry on the drive axle may need to be more aggressive than
anti-squat geometry.
▪ Swing arm length and angle dictates the rate of change of geometry forces.
▪ For an independent suspension a percentage of 100% would indicate the
suspension is taking 100% of the load transfer under acceleration/braking instead of
the springs which effectively binds the suspension.
▪ However, in the case of leaf spring rear suspension the anti-squat can often exceed
100% (meaning the rear may actually raise under acceleration) and because there
isn't a second arm to bind against, the suspension can move freely.
▪ Traction bars are often added to drag racing cars with rear leaf springs to increase
the anti-squat to its maximum. This has the effect of forcing the rear of the car
upwards and the tires down onto the ground for better traction.
61

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).
62

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

Vehicle load transfer part II 2021

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