Vehicle Load Transfer PartI_II_23MAR13

7/29/2019 Vehicle Load Transfer PartI_II_23MAR13

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1

Wm Harbin

Technical Director

BND TechSource

Vehicle Load Transfer


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Part I

General Load Transfer


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Within any modern vehicle suspension there are manyfactors

to consider during design and development.

Factors in vehicle dynamics:

Vehicle ConfigurationVehicle Type (i.e. 2 dr Coupe, 4dr Sedan, Minivan, Truck, etc.) Vehicle Architecture (i.e. FWD vs. RWD, 2WD vs.4WD, etc.)

Chassis Architecture (i.e. type: tubular, monocoque, etc. ; material: steel, aluminum,carbon fiber, etc. ; fabrication: welding, stamping, forming, etc.)

Front Suspension System Type (i.e. MacPherson strut, SLA Double Wishbone, etc.)

Type of Steering Actuator (i.e. Rack and Pinion vs. Recirculating Ball) Type of Braking System (i.e. Disc (front & rear) vs. Disc (front) & Drum (rear))

Rear Suspension System Type (i.e. Beam Axle, Multi-link, Solid Axle, etc.)

Suspension/Braking Control Systems (i.e. ABS, Electronic Stability Control,Electronic Damping Control, etc.)

Factors in Vehicle Dynamics


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Factors in vehicle dynamics (continued):

Vehicle Suspension GeometryVehicle Wheelbase Vehicle Track Width Front and Rear

Wheels and Tires

Vehicle Weight and Distribution

Vehicle Center of Gravity Sprung and Unsprung Weight

Springs Motion Ratio

Chassis Ride Height and Static Deflection

Turning Circle or Turning Radius (Ackermann Steering Geometry)

Suspension Jounce and Rebound

Vehicle Suspension Hard Points: Front Suspension

Scrub (Pivot) Radius

Steering (Kingpin) Inclination Angle (SAI) Caster Angle

Mechanical (or caster) trail

Toe Angle

Camber Angle

Ball Joint Pivot Points

Control Arm Chassis Attachment Points

Knuckle/Brakes/Steering

Springs/Shock Absorbers/Struts

ARB (anti-roll bar)



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Vehicle Suspension Geometry (continued)Vehicle Suspension Hard Points (continued):

Rear Suspension Scrub (Pivot) Radius

Steering (Kingpin) Inclination Angle (SAI)

Caster Angle (if applicable)

Mechanical (or caster) trail (if applicable)

Toe Angle

Camber Angle Knuckle and Chassis Attachment Points

Various links and arms depend upon the Rear Suspension configuration.

(i.e. Dependent vs. Semi-Independent vs. Independent Suspension)

Knuckle/Brakes

Springs/Shock Absorbers

ARB (anti-roll bar)

Vehicle Dynamic Considerations

Suspension Dynamic Targets Wheel Frequency

Bushing Compliance

Lateral Load Transfer with and w/o ARB

Roll moment

Roll Stiffness (degrees per g of lateral acceleration)

Maximum Steady State lateral acceleration (in understeer mode)

Rollover Threshold (lateral g load)

Linear Range Understeer (typically between 0g and 0.4g)



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Vehicle Dynamic Considerations (continued) Suspension Dynamic Analysis

Bundorf Analysis

Slip angles (degrees per lateral force) Tire Cornering Coefficient (lateral force as a percent of rated vertical load per degree slip

angle)

Tire Cornering Forces (lateral cornering force as a function of slip angle)

Linear Range Understeer

Steering Analysis Bump Steer Analysis

Roll Steer Analysis

Tractive Force Steer Analysis

Brake Force Steer Analysis

Ackerman change with steering angle

Roll Analysis Camber gain in roll (front & rear)

Caster gain in roll (front & rear if applicable)

Roll Axis Analysis

Roll Center Height Analysis

Instantaneous Center Analysis

Track Analysis

Load Transfer Analysis Unsprung and Sprung weight transfer

Jacking Forces

Roll Couple Percentage Analysis

Total Lateral Load Transfer Distribution (TLLTD)



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While the total amount of factors may seem a bit overwhelming, it may be

easier to digest if we break it down into certain aspects of the total.

The intent of this document is to give the reader a better understanding of

vehicle dynamic longitudinal and lateral load transfer as a vehicle

accelerates/decelerates in a particular direction.

The discussion will include:

Part I General Load Transfer Information Load vs. Weight Transfer

Rotational Moments of Inertia

Sprung and Unsprung Weight

Part II Longitudinal Load Transfer Vehicle Center of Gravity Longitudinal Load Transfer

Suspension Geometry Instant Centers

Side View Swing Arm

Anti-squat, Anti-dive, and Anti-lift


Part III Lateral Load Transfer Cornering Forces

Suspension Geometry

Front View Swing Arm Roll Center Heights

Roll Axis

Roll Stiffness

Anti-roll bars Tire Rates

Roll Gradient

Lateral Load Transfer


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Load vs. Weight Transfer


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In automobiles, load transfer is the imaginary "shifting" of

weight around a motor vehicle during acceleration (both

longitudinal and lateral). This includes braking, or deceleration

(which can be viewed as acceleration at a negative rate). Load

transfer is a crucial concept in understanding vehicle dynamics.

Often load transfer is misguidedly referred to as weight

transfer due to their close relationship. The difference being

load transfer is an imaginary shift in weight due to an

imbalance of forces, while weight transfer involves the actual

movement of the vehicles center of gravity (Cg). Both result in

a redistribution of the total vehicle load between the

individual tires.



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Weight transfer involves the actual (small) movement of the

vehicle Cg relative to the wheel axes due to displacement of

liquids within the vehicle, which results in a redistribution of

the total vehicle load between the individual tires.

Liquids, such as fuel, readily flow within their containers,

causing changes in the vehicle's Cg. As fuel is consumed, not

only does the position of the Cg change, but the total weight

of the vehicle is also reduced.

Another factor that changes the vehicles Cg is the expansion

of the tires during rotation. This is called dynamic rolling

radius and is effected by wheel-speed, temperature, inflation

pressure, tire compound, and tire construction. It raises the

vehicles Cg slightly as the wheel-speed increases.



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The major forces that accelerate a vehicle occur at the tires

contact patch. Since these forces are not directed through the

vehicle's Cg, one or more moments are generated. It is these

moments that cause variation in the load distributed between

the tires.

Lowering the Cg towards the ground is one method of

reducing load transfer. As a result load transfer is reduced in

both the longitudinal and lateral directions. Another method

of reducing load transfer is by increasing the wheel spacings.

Increasing the vehicles wheel base (length) reduces

longitudinal load transfer. While increasing the vehicles track

(width) reduces lateral load transfer.



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Rotational Moments of Inertia

12

y

x

zVertical

Lateral

Longitudinal

Roll

(p)

Yaw

(r) Pitch

(q)

Cg


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Moment of Inertia

Polar moment of inertia A simple demonstration of polar moment of inertia is to compare a

dumbbell vs. a barbell both at the same weight. Hold each in the

middle and twist to feel the force reacting at the center. Notice the

dumbbell (which has a lower polar moment) reacts quickly and the

barbell (which has a higher polar moment) reacts slowly.

13

Wdo2=

d1 d1

CL

d2 d2

CL

W W

Example:

W = 50 lb (25 lbat each end)

d1 = 8 in

d2 = 30 in

22

1 3200)8(*25*2 inlb ==22

2 000,45)30(*25*2 inlb ==


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Moment of Inertia

Sum the polar moments of inertia The total polar moment of inertia for a vehicle can be

determined by multiplying the weight of each component by

the distance from the component Cg to the Cg of the vehicle.

The sum of the component polar moments of inertia would

establish the total vehicle polar moment of inertia.

A vehicle with most of its weight near the vehicle Cg has a

lower total polar moment of inertia is quicker to respond to

steering inputs.

A vehicle with a high polar moment is slower to react to

steering inputs and is therefore more stable at high speed

straight line driving.

14


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Moment of Inertia

Effects of polar moments of inertia

Here is an example of a V8 engine with a typical transmissionpackaged into a sports car.

15

Example:

WEng = 600 lb

WTran = 240 lb

dEng = 40 indTran = 10 in

dEng

dTran

dWdWM TranTranEngEngo22 )()()( +=

222 000,984)10(240)40(600)( inlbinlbinlbMo

=+=


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Moment of Inertia

16

Example:

WEng = 600 lb

WTran = 240 lb

dEng = 70 indTran = 40 in

dEng

dTran

Effects of polar moments of inertia

Here is an example of a V8 engine with a typical transmissionpackaged into a sedan.

dWdWM TranTranEngEngo

22 )()()( +=

222 000,324,3)40(240)70(600)( inlbinlbinlbM o =+=


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Load Transfer


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Load Transfer

Load Transfer

The forces that enable a road vehicle to accelerate and stop

all act at the road surface.

The center of gravity, which is located considerably above

the road surface, and which is acted upon by the

accelerations resulting from the longitudinal forces at the

tire patches, generates a moment which transfers load.

As asymmetric load results in differing traction limits, avehicles handling is affected by the dynamic load

distribution.


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Load Transfer equations & terms

Load Transfer

g

onAcceleratiVehicleWeightVehicleForceInertial

*=

Wheelbase

CGForceInertialTransferLoad

height*=

am=FLawSecondsNewton :'

F = forcem = mass

a = acceleration

g = ag = acceleration due to gravity

= 32.2ft/sec2 = 9.8m/sec2

ax = acceleration in the x directionay = acceleration in the y direction

az = acceleration in the z direction

Weight = mass * ag


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Load Transfer

Load transfers between the Center of Gravity and the roadsurface through a variety of paths.

Suspension Geometry

Front: Location of instant centers

(Side View Swing Arm) Rear: Instant centers, Lift Bars

(Side View Swing Arm)

Suspension Springs

Front: Coils, Air Springs, leafs or Torsion bars and Anti-roll bars

Rear: Coils, Air Springs, leafs or Torsion bars and Anti-

roll bars


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Load transfer (continued)

Dampers (Shock Absorbers)

During transient conditions

Tires

During all conditions (where the rubber meets the road)

Where and how you balance the load transfer between the

Springs, Geometry, Dampers and Tires are key determinates as

to how well the car will accelerate and brake and the stability

associated with each condition.

Load Transfer


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Load Transfer Control Devices

Dampers (Shock Absorbers)

Along with the springs, dampers transfer the load of the rolling

(pitching) component of the vehicle. They determine how the

load is transferred to and from the individual wheels while the

chassis is rolling and/or pitching.

Within 65-70% critically damped is said to be the ideal damper

setting for both handling and comfort simultaneously. Most

modern dampers show some digression to them as well,

meaning they may be 70% critically damped at low piston

speeds but move lower to allow the absorption of large bumps.Damping is most important below 4 in/second as this is where

car control tuning takes place.


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Springs Along with the dampers (shock absorbers), springs transfer the load

of the sprung mass of the car to the road surface. Duringmaneuvers, depending on instant center locations, the springs anddampers transfer some portion of the (m x a), mass x acceleration,forces to the ground.

Spring Rate is force per unit displacement for a suspension springalone .

For coil springs this is measured axially along the centerline.

For torsion bar springs it is measured at the attachment arm.

For leaf springs it is measured at the axle seat.

The spring rate may be linear (force increases proportionally with

displacement) or nonlinear (increasing or decreasing rate with

increasing displacement).

Units are typically lb/in.


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Anti-roll bars [Drawing 1] shows how an anti-roll bar (ARB) is twisted whenthe body rolls in a turn. This creates forces at the four pointswhere the bar is attached to the vehicle. The forces are shownin [Drawing 2]. Forces A on the suspension increase [load]transfer to the outside tire. Forces B on the frame resist bodyroll. The effect is a reduction of body roll and an increase in

[load] transfer at the end of the chassis which has the anti-rollbar. Because the total [load] transfer due to centripetal force isnot changed, the opposite end of the chassis has reduced [load]transfer. [6]

Drawing 2 Drawing 1Direction of Turn

A A

BB


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Bushing Deflection (suspension compliance) All of the calculations shown in this presentation do not include

bushing deflection. There are many rubber bushings within a vehiclesuspension to consider when analyzing suspension compliance.



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Frame/Chassis Deflection All of the calculations shown in this presentation are made under

the assumption that the frame or chassis is completely rigid (both intorsion and bending). Of course any flexing within the frame/chassis

will adversely effect the performance of the suspension which is

attached to it.


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100% Unsprung weight includes the mass of the tires, rims,brake rotors, brake calipers, knuckle assemblies, and ball

joints which move in unison with the wheels.

50% unsprung and 50% sprung weight would be comprisedof the linkages of the wheel assembly to the chassis.

The % unsprung weight of the shocks, springs and anti-roll

bar ends would be a function of their motion ratio/2 with

the remainder as % sprung weight.

The rest of the mass is on the vehicle side of the springs is

suspended and is 100% sprung weight.

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The shocks, springs, struts and anti-roll bars are normally mounted at

some angle from the suspension to the chassis. Motion Ratio: If you were to move the wheel 1 inch and the spring were

to deflect 0.75 inches then the motion ratio would be 0.75 in/in.

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Motion Ratio = (B/A) * sin(spring angle)


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The shocks, springs, struts and anti-roll bars are normally mounted at

some angle from the suspension to the chassis.

Motion Ratio: If you were to move the wheel 1 inch and the spring were

to deflect 0.75 inches then the motion ratio would be 0.75 in/in.

Motion Ratio =

(B/A) * sin(spring angle)


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Wheel Rates

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Wheel Rates are calculated by taking the square of the motion ratio times

the spring rate. Squaring the ratio is because the ratio has two effects on

the wheel rate. The ratio applies to both the force and distance traveled.

Because it's a force, and the lever arm is multiplied twice.

The motion ratio is factored once to account for the distance-traveled differentialof the two points (A and B in the example below).

Then the motion ratio is factored again to account for the lever-arm forcedifferential.

Example: K

|

A----B------------P

P is the pivot point, B is the spring mount, and A is the wheel. Here the motion

ration (MR) is 0.75... imagine a spring K that is rated at 100 lb/in placed at Bperpendicular to the line AP. If you want to move A 1 in vertically upward, B would

only move (1in)(MR) = 0.75 in. Since K is 100 lb/in, and B has only moved 0.75 in,

there's a force at B of 75 lb. If you balance the moments about P, you get

75(B)=X(A), and we know B = 0.75A, so you get 75(0.75A) = X(A). A's cancel and you

get X=75(0.75)=56.25. Which is [100(MR)](MR) or 100(MR)2.

Wheel Rate (lb/in) =

(Motion Ratio)2* (Spring Rate)


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0

50

100

150

200

250

300

-2.49 -2.22 -1.95 -1.68 -1.42 -1.15 -0.89 -0.63 -0.36 -0.10 0.16 0.41 0.67 0.93 1.18 1.44 1.69 1.94 2.19 2.44

WheelRate(lb/in

)

Rebound to Jounce (in)

Wheel Rate vs. Wheel Position

Coil-over Shock

ARB

Wheel Rates

32

Since the linkages pivot, the spring angles change as the components

swing along an arc path. This causes the motion ratio to be calculated

through a range. The graph below shows an example of these results forboth coil-over shock and anti-roll bar for an independent front suspension

from rebound to jounce positions.

KW = Wheel Rate (lb/in) =

(Motion Ratio - range)2* (Spring Rate - linear)

KW = MR2 * KS

Example:

Coil-over KS = 400 lb/in (linear)Coil-over MR

= 0.72-.079 in/in

ARB KS = 451.8 lb/in (body roll)ARB MR= 0.56-0.61 in/in

Ride height


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0

50

100

150

200

250

300

-2.49 -2.22 -1.95 -1.68 -1.42 -1.15 -0.89 -0.63 -0.36 -0.10 0.16 0.41 0.67 0.93 1.18 1.44 1.69 1.94 2.19 2.44

WheelRate(lb/in

)

Rebound to Jounce (in)

Wheel Rate vs. Wheel Position

Coil-over Shock

ARB

Wheel Rates

33

In longitudinal pitch, the anti-roll bar (ARB) rotates evenly as the chassis

moves relative to the suspension. Therefore, the ARB only comes into play

during lateral pitch (body roll) of the vehicle (it also comes into play duringone wheel bump, but that rate is not shown here).

KW = Wheel Rate (lb/in) =

(Motion Ratio - range)2* (Spring Rate - linear)

KW = MR2 * KS

Example:

Coil-over KS = 400 lb/in (linear)Coil-over MR= 0.72-.079 in/in

ARB KS = 451.8 lb/in (body roll)ARB MR= 0.56-0.61 in/in

Ride height


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Spring Rates/Ride Frequency

34

The static deflection of the suspension determines its natural frequency.

Static deflection is the rate at which the suspension compresses in

response to weight.

0

20

40

60

80

100

120

140

160

180

200

0 10 20 30 40 50 60 70 80 90 100

w=

Frequency(cycles/min)

x = Static Deflection (in)

Ride Natural Frequency vs. Static Wheel Deflection

x

188=


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Ride frequency is the undamped natural frequency of the body in ride.

The higher the frequency, the stiffer the ride.

Based on the application, there are ballpark numbers to consider.

30 - 70 CPM for passenger cars

70 - 120 CPM for high-performance sports cars

120 - 300+ CPM for high downforce race cars

It is common to run a spring frequency higher in the rear than the front.

The idea is to have the oscillation of the front suspension finish at the

same time as the rear.

Since the delay between when the front suspension hits a bump and the

rear suspension hits that bump varies according to vehicle speed, the

spring frequency increase in the rear also varies according to the

particular speed one wants to optimize for.


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Once the motion ratios has been established, the front and rear spring

rates can be optimized for a flat ride at a particular speed.


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Part II

Longitudinal Load Transfer


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Center of Gravity


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Locating the center of gravity in the X-Y (horizontal) plane isperformed 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.

Center of Gravity Location


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WR

lRFL

Cg

WF

SCALE SCALE

Center of Gravity

Horizontal Plane Location


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100%

+=

tot

LFRFfront

W

WWW

100%

+=

tot

RRLR

rear W

WW

W

100%

+=

tot

LRLFleft

W

WWW 100%

+=

tot

RRRFright

W

WWW

100%

+=

tot

LRRFdiag

W

WWW


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42


1003542

880880%7.49

+== frontW

1003542

891891%3.50

+== rearW

1003542

891880%50

+== leftW

1003542

891880%50

+== rightW

1003542

891880%50

+== diagW

Example: C3 Corvette Upgrade

Weight total = 3542 lbW RF = 880 lb

W LF = 880 lb

W RR = 891 lb

W LR = 891 lb


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LW

W

LW

W

tot

r

tot

f

f=

= 1

LW

WL

W

W

tot

f

tot

rr =

= 1


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44


983542

178298

3542

17601in3.49 =

==

f

983542

176098

3542

17821in7.48 =

== r


W tot = 3542 lbW F = 1760 lb

W R = 1782 lb

L = 98.0 in


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Center of Gravity Height Location

== LWWLMHorizontal fRR 0:

( )

coscoscossin0

coscossin0:

LWLWLWhWor

LWWLWhWMRaised

ffRcg

ffRcgR

+=

++==


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The center of gravity height above the spindle centerline is:

hW

LW

W

LW

abovehtot

f

tot

f

cg

=

= *

*

sin*

cos**2

cos

sintan =

=

L

h

hCGtotal on level ground is:

(where rt = tire radius)t

tot

f

cg rhW

LWtotalh +

=

*

*2


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The center of gravity height

above the spindle centerline is:

12*3542

9604*62.22

98.6sin*3542

98.6cos*98*62.22

11.5 ==

98.6cos

98.6sin

98

1298.6tan ==

hCGtotal on level ground is:

(where rt = tire radius)89.11

12*3542

9604*62.22in17 +=

Example: C3 Corvette

W tot = 3542 lbW F = 1760 lb

W R = 1782 lb

L = 98.0 in

rt = 11.89 in

Dh = 12 inDWF= 22.62 lb


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Longitudinal (vehicle fore-aft direction) load transfer occursdue 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.


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The vehicle load distribution on level ground is shown in the

following equations.

A B

WF WR

lRFL

Cg

WT

=

LW=WW=LWAxleFront

RFRF

= LW=WW=LWAxleRear

FRFR


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The vehicle load distribution on level ground is shown in the

following equations.

A B

WF WR

lRFL

Cg

WT

lb=WW=LWAxleFront FRF 176098

7.483542 =

=

lb=WW=LWAxleRear RFR 178298

3.493542 =

=


W T = 3542 lbW F = 1760 lb

W R = 1782 lb

L = 98.0 in

lF = 49.3 inlR = 48.7 in


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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

lRFL

acceleration

force

Cg

accel FaWT

hcg

Lh

aaW

Lham=

LhF=W cg

g

xT

cgcgaT **=


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53


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

lRFL

acceleration

force

Cgaccel

FaWT

hcg

lb=LhF=W cgaT 2.286

9817*)2.32/15(*3542 =


W T = 3542 lb

L = 98.0 inhcg = 17 in

ax = 15 ft/sec2

ag = 32.2 ft/sec2


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54

L

h

a

a

LW=WW=0=M(B)

cg

g

xRTF *


In forward acceleration the load on the front axle can befound by solving the moment about point B of the figure.

A B

WF - W WR + W

lRFL

accelerationforce

Cgaccel FaWT

hcg


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55


In forward acceleration the load on the front axle can befound by solving the moment about point B of the figure.


W T = 3542 lbW F = 1760 lbDW = 286.2 lblR = 48.7 inL = 98.0 in

hcg = 17 in

ax = 15 ft/sec2

ag = 32.2 ft/sec2

A B

WF - W WR + W

lRFL

acceleration

force

Cgaccel

FaWT

hcg

lb=WW=0=M(B) F 8.147398

17*

2.32

15

98

7.483542 =

Check: 1473.8 lbs + 286.2 lbs = 1760 lb


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56


In forward acceleration the load on the rear axle can befound by solving the moment about point A of the figure.

A B

WF + W WR - W

lRFL

accelerationforce

Cgaccel FaWT

hcg

++

L

h

a

a

LW=WW=0=M(A)

cg

g

xFTR *


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57


In forward acceleration the load on the rear axle can befound by solving the moment about point A of the figure.

A B

WF + W WR - W

lRFL

acceleration

force

Cgaccel Fa

WT

hcg


W T = 3542 lbW R = 1782 lbDW = 286.2 lblF = 49.3 inL = 98.0 in

hcg = 17 in

ax = 15 ft/sec2

ag = 32.2 ft/sec2

lbs=WW=0=M(A) R 2.206898

17*

2.32

15

98

3.493542 =

++

Check: 2068.2 lbs - 286.2 lbs = 1782 lbs


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58

Longitudinal Acceleration Pitch

The vehicle pitch angle , if no suspension forces oppose it (noanti-dive, anti-lift, or anti-squat), is a function of the load

transfer (W) and the wheel rate (KW).

If the front wheel rate is KWf and the rear is KWr then the load

transfer (W) would be divided between the left and rightwheels. 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)


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59


KW=Z

KW=Z

Wf

F

Wr

R 2/2/

180

L

K

W+

K

W

=L

Z+Z

=L

K

W+

K

W

=

WrWf

deg

FRWrWf

rad *

2/2/2/2/

ZR ZF

Cg Cg

(no anti-dive, anti-lift, or anti-squat)



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60


in=Zin=Z FR 62.26.230

2/2.286455.

35.314

2/2.286==

deg63.*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

radrad


KSf= 400 lb/in MRf= 0.759 in/in KWf= 230.26 lb/inKSr = 600 lb/in MRr = 0.724 in/in KWr = 314.35 lb/inDW = 286.2 lbL = 98.0 in

ZR ZF

Cg Cg


KW = MR2 * KS


61/101

61

Suspension Geometry


62/101

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.

62

Front View GeometrySide View Geometry


63/101

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 pointsattaching 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.

63FVSA

Swing Arm

SVSA

Swing Arm


64/101

Swing Arms

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.

64

FVSA

Swing Arm

SVSA

Swing Arm

id i i


65/101

65

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 anddifferential move together and are suspended from the chassis

using springs and/or trailing arms.

Side View Swing Arm

Cg

e

d

L

Fza

Fxa

Fx

Fz

IC

h

Sid i S i


66/101

66

Shown is a schematic of an independent rear suspension (IRS)

with the linkages replaced with a swing arm. In an IRS thedifferential is mounted to the chassis as the half shafts move

independently and are suspended from the chassis using

springs, control arms and trailing arms.

Side View Swing Arm

Cg

e

d

L

Fza

Fxa

Fx

Fz

IC

h

b

r


67/101

67

Anti- Geometry


68/101

68

Anti- Geometry

Anti-squat

Anti-squat in rear suspensions reduces the jounce (upward)

travel during forward acceleration on rear wheel drive cars only.

Anti-diveAnti-dive geometry in front suspensions reduces the jounce

(upward) deflection under forward braking.

Anti-liftAnti-lift in rear suspensions reduces rebound (downward) travel

during forward braking.


69/101

69

Anti-Squat Geometry


70/101

70

Anti-squat geometry

Anti-squatDuring 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


71/101

71

Anti-squat geometry

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

br

WT

100% Anti-squat

Line


72/101

72

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.

Anti-squat geometry


73/101

73

Anti-squat geometry

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).

Lh=

dre= tan

Cg

e

d

L

Fza

Fxa

Fx

Fz

IC

h

b

r

WT

100% Anti-squat

Line


74/101

74

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.

Anti-squat geometry


75/101

75

The load transfer during acceleration is as shown:

The tractive effort (Fx) at the drive wheels is calculated as:

L

ha

a

W=

L

ham=F x

g

T

za

x

g

T

xxa aaW=F=F

Anti-squat geometry

(1)

(2)


76/101

76

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

reF=F

0=reF-dF0=M

xaza

xazaR

d

rea

a

W=F x

g

T

za

Anti-squat geometry

(3)

(4)

Cg

e

d

L

Fza

Fxa

Fx

Fz

IC

h

b

rWT

100% Anti-squat

Line


77/101

77

Equating equations (1) and (4) results in an equation whichindicates the relationships for 100% anti-squat.

The angle the instant center (IC) must lie on for 100% anti-

squat is:

d

re=

L

h

d

rea

a

W=

L

ha

a

W=F x

g

T

x

g

T

za

L

h=

d

re=

tan

Anti-squat geometry

(5)

(6)


78/101

78

If the tan < h/L then squat will occur.

L

h=

d

re=

tan

Anti-squat geometry

Cg

e

d

L

Fza

Fxa

Fx

Fz

IC

h

b

rWT

100% Anti-squat

Line

Anti squat geometry


79/101

79

Anti-squat rear solid axle

Shown is a SVSA of a solid rear drive axle. The torque reaction istaken by the suspension components.

Anti-squat geometry

Cg

e

d

L

Fza

Fxa

Fx

Fz

IC

h

L

h

d

e=squatAnti =%100

100*/

/

/

tan%

Lh

de

Lh=squatAnti =

Anti squat geometry


80/101

80

Anti-squat rear solid axle

Shown is a SVSA of a solid rear drive axle. The torque reaction istaken by the suspension components.

Anti-squat geometry

Cg

e

d

L

Fza

Fxa

Fx

Fz

IC

h

%125100*100/20

45/25.11

/

tan% ==

Lh=squatAnti

L

h

d

e=squatAnti =%100

Example: Solid Axlee= 11.25 ind = 45 in

L = 100 in

h = 20 in

Anti squat geometry


81/101

81

Anti-squat independent rear suspension

Shown is a SVSA of an independent rear suspension (IRS). Thetorque reaction is taken by the chassis.

Anti-squat geometry

L

h

d

re=squatAnti =

%100

100*/

/

/

tan%

Lh

dre

Lh=squatAnti

=

Cg

e

d

L

Fza

Fxa

Fx

Fz

IC

h

b

rWT

100% Anti-squat

Line

Anti squat geometry


82/101

82

Anti-squat independent rear suspension

Shown is a SVSA of an independent rear suspension (IRS). Thetorque reaction is taken by the chassis.

Anti-squat geometry

L

h

d

re=squatAnti =

%100

Example: C3 Upgradee= 15.62 ind = 32.84 in

L = 98 in

h = 17 in

r = 11.89 in

%5.65100*98/17

84.32/89.1162.15

/

tan% =

=

Lh=squatAnti

Cg

e

d

L

Fza

Fxa

Fx

Fz

IC

h

b

rWT

100% Anti-squat

Line


83/101

83

Anti-squat geometry

Anti-squat effects on longitudinal pitch angle

100*%

*

=oncompensatipitch

Wf

Wr

K

K

L

h

L

h

d

re

+

+

Lh

Kdre

KLh

K WfWrWrx

g

Ta

a

W

L=(rad)anglePitch *

2*1*

2*1*

2*1**

1

Anti-squat geometry


84/101

84

q g y


Example: C3 Corvette Upgradee r = 3.73 in WT = 3542 lbd = 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

+ 9817*

)2*26.230(1

84.3273.3*

)2*35.314(1

9817*

)2*35.314(1180*

4.3863542*

981

=(rad)anglePitch

squatAntidegrad=(deg)anglePitch %[email protected]

*0079.0 =

%28100*4103.1136.% ==oncompensatipitch

Check: (1-.28)*.63deg = .45deg

Anti-squat geometry


85/101

85

q g y


Example: C3 Corvette Upgradee r = 3.73 in WT = 3542 lbd = 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

0*)2*35.314(

1

98

17

*)2*35.314(

1

180*4.386

3542

*98

1

=(rad)anglePitch

squatAntidegrad=(deg)anglePitch %[email protected]

*011.0 =

If e-r/d = 0, then Anti-squat = 0



86/101

86

g

in=Zin=Z FR 62.26.230

2/2.286455.

35.314

2/2.286==

deg63.*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

radrad


KSf= 400 lb/in MRf= 0.759 in/in KWf= 230.26 lb/inKSr = 600 lb/in MRr = 0.724 in/in KWr = 314.35 lb/inDW = 286.2 lbL = 98.0 in

ZR ZF

Cg Cg


KW = MR2 * KS

Check: if e-r/d = 0,

then anti-squat = 0

0.63 @ 0% Anti-squat

Previous Slide

No. 60


87/101

87

Anti-dive Geometry

Anti-dive geometry


88/101

88

g y

Anti-diveDuring 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.

Cg

JOUNCE

REBOUND

Anti-dive geometry


89/101

89

g y Anti-dive

The total longitudinal load transfer under steady acceleration or

braking is a function of the wheelbase (L), CG height (h), andbraking force (WT)*(ax/ag).

Cg

L

h

WT

WT (ax/ag) = braking force

+ load- load

L

ha

a

W=

L

ham=load x

g

T

Anti-dive geometry


90/101

90

g y Anti-dive

The total longitudinal load transfer under steady acceleration or

braking is a function of the wheelbase (L), CG height (h), andbraking force (WT)*(ax/ag).

lb=

L

ham=load 2.286

98

17*180*

4.386

3542=


WT = 3542 lbL = 98 in

h = 17 inax = 180 in/sec

2 (.46 g)

ag = 386.4 in/sec2

Anti-dive geometry


91/101

91

g y

Example: C3 Corvette UpgradeWT = 3542 lbL = 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

AxleLoad(lbf)

Deceleration (g's)

Load Transfer vs. Deceleration

Front Load (lbf)Rear Load (lbf)

L

ha

a

W=

L

ham=load x

g

T

Anti-dive geometry


92/101

92

g y

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 transferredto the tire contact patch and reacted against the ground:

Rotor effective diameter

Caliper piston diameter

Lining friction coefficients

Tire traction coefficient properties

Cg

JOUNCE

REBOUND

Anti-dive geometry


93/101

93

g y


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.

Cg

JOUNCE

REBOUND

Anti-dive geometry


94/101

94

g y


Cg

JOUNCE

REBOUND

0%

10%

20%

30%

40%

50%

60%

70%

80%

90%

100%

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

%ofVehicleLoadTransfer

%ofTotalBrakingForce

Deceleration (g's)

Typical Brake System Bias

% Front Load

% Front Braking

% Rear Load

% Rear Braking

Anti-dive geometry


95/101

95

g y

100*)(%*)/(*tan% brakingfronthL=diveAnti f

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 ICs are below these lines, the percentage of anti will bebelow 100%. If the ICs are above these lines, the percentage of anti will be above

100%.

Cg

L

ICr

h

brWT

ax

bfICf%FB x L

100% Anti-dive

Line

100% Anti-lift

Line

1 - %FB x L

100*)%1(*)/(*tan% brakingfronthL=liftAnti r

Anti-dive geometry


96/101

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 ICs are below these lines, the percentage of anti will bebelow 100%. If the ICs are above these lines, the percentage of anti will be above

100%.

96


tanbf= .1124tanb

r= .4894

L = 98 in

h = 17 in

% front braking @ .46g = .71

%46100*)71(.*)765.5(*1124.% ==diveAnti

%82100*)29(.*)765.5(*4894.% ==liftAnti

Anti-dive geometry


97/101

Anti-dive (front) and Anti-lift (rear) suspension

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 rateof deceleration changes.

97

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

%

ofTotalBrakingForce

Deceleration (g's)

Anti-dive & Anti-lift vs. Brake Bias

% Anti-dive

% Front Braking

% Anti-lift

% Rear Braking

Design factors in


98/101

98

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.

Design factors in


99/101

99

es g facto s

Anti-dive and Anti-Squat

For an independent suspension a percentage of 100% wouldindicate 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-squatcan 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.

References:


100/101

100

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 AutomotiveEngineers 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).


101/101

Next

Part III - Lateral Load TransferThank You!

For additional information please visit our free website at:

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