vehicle longitudinal load transfer parts i_ii_mar13

1

Wm Harbin

Technical Director

BND TechSource

Vehicle Load Transfer

2


Part I

General Load Transfer

3

Within any modern vehicle suspension there are many factors to consider during design and development.

Factors in vehicle dynamics: • Vehicle Configuration

• Vehicle Type (i.e. 2 dr Coupe, 4dr Sedan, Minivan, Truck, etc.)

• Vehicle Architecture (i.e. FWD vs. RWD, 2WD vs.4WD, etc.)

• Chassis Architecture (i.e. type: tubular, monocoque, etc. ; material: steel, aluminum,

carbon fiber, etc. ; fabrication: welding, stamping, forming, etc.)

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

• Type of Steering Actuator (i.e. Rack and Pinion vs. Recirculating Ball)

• Type of Braking System (i.e. Disc (front & rear) vs. Disc (front) & Drum (rear))

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

• Suspension/Braking Control Systems (i.e. ABS, Electronic Stability Control,

Electronic Damping Control, etc.)

Factors in Vehicle Dynamics

4

Factors in vehicle dynamics (continued): • Vehicle Suspension Geometry

• Vehicle Wheelbase • Vehicle Track Width Front and Rear • Wheels and Tires

• Vehicle Weight and Distribution • Vehicle Center of Gravity

• Sprung and Unsprung Weight • Springs Motion Ratio

• Chassis Ride Height and Static Deflection • Turning Circle or Turning Radius (Ackermann Steering Geometry)

• Suspension Jounce and Rebound • Vehicle Suspension Hard Points:

• Front Suspension • Scrub (Pivot) Radius • Steering (Kingpin) Inclination Angle (SAI) • Caster Angle • Mechanical (or caster) trail • Toe Angle • Camber Angle • Ball Joint Pivot Points • Control Arm Chassis Attachment Points • Knuckle/Brakes/Steering • Springs/Shock Absorbers/Struts • ARB (anti-roll bar)


5

Factors in vehicle dynamics (continued): • Vehicle Suspension Geometry (continued)

•Vehicle Suspension Hard Points (continued): • Rear Suspension

• Scrub (Pivot) Radius • Steering (Kingpin) Inclination Angle (SAI) • Caster Angle (if applicable) • Mechanical (or caster) trail (if applicable) • Toe Angle • Camber Angle • Knuckle and Chassis Attachment Points

• Various links and arms depend upon the Rear Suspension configuration. (i.e. Dependent vs. Semi-Independent vs. Independent Suspension)

• Knuckle/Brakes • Springs/Shock Absorbers • ARB (anti-roll bar)

•Vehicle Dynamic Considerations • Suspension Dynamic Targets

• Wheel Frequency • Bushing Compliance • Lateral Load Transfer with and w/o ARB • Roll moment • Roll Stiffness (degrees per g of lateral acceleration)

• Maximum Steady State lateral acceleration (in understeer mode)

• Rollover Threshold (lateral g load)

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


6

Factors in vehicle dynamics (continued): • Vehicle Dynamic Considerations (continued)

• Suspension Dynamic Analysis • Bundorf Analysis

• Slip angles (degrees per lateral force) • Tire Cornering Coefficient (lateral force as a percent of rated vertical load per degree slip angle) • Tire Cornering Forces (lateral cornering force as a function of slip angle) • Linear Range Understeer

• Steering Analysis • Bump Steer Analysis • Roll Steer Analysis • Tractive Force Steer Analysis • Brake Force Steer Analysis • Ackerman change with steering angle

• Roll Analysis • Camber gain in roll (front & rear) • Caster gain in roll (front & rear – if applicable) • Roll Axis Analysis • Roll Center Height Analysis • Instantaneous Center Analysis • Track Analysis

• Load Transfer Analysis • Unsprung and Sprung weight transfer • Jacking Forces

• Roll Couple Percentage Analysis • Total Lateral Load Transfer Distribution (TLLTD)


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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

9

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 vehicle’s Cg is the expansion of the tires during rotation. This is called “dynamic rolling radius” and is effected by wheel-speed, temperature, inflation pressure, tire compound, and tire construction. It raises the vehicle’s Cg slightly as the wheel-speed increases.


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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.


Rotational Moments of Inertia

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y

x

z Vertical

Lateral

Longitudinal

Roll (p)

Yaw (r) Pitch

(q)

Cg

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.

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Wd o2

d1 d1

C L

d2 d2

C L

W W

Example:

W = 50 lb (25 lb

at each end)

d1 = 8 in

d2 = 30 in

221 3200)8(*25*2 inlb 22

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

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.

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Moment of Inertia

Effects of polar moments of inertia • Here is an example of a V8 engine with a typical transmission

packaged into a sports car.

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Example: WEng = 600 lb WTran = 240 lb dEng = 40 in dTran = 10 in

dEng

dTran

dW dWM TranTranEngEngo22 )()()(

222 000,984)10(240)40(600)( inlb inlb inlbM o

Moment of Inertia

16

Example: WEng = 600 lb WTran = 240 lb dEng = 70 in dTran = 40 in

dEng

dTran

Effects of polar moments of inertia • Here is an example of a V8 engine with a typical transmission

packaged into a sedan.

dW dWM TranTranEngEngo22 )()()(

222 000,324,3)40(240)70(600)( inlb inlb inlbM 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, a vehicles handling is affected by the “dynamic load distribution”.

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Load Transfer equations & terms

Load Transfer

g

onAcceleratiVehicleWeightVehicleForceInertial

*

Wheelbase

CGForceInertialTransferLoad

height*

a m = FLawSecondsNewton :'

F = force

m = mass

a = acceleration

g = ag = acceleration due to gravity

= 32.2ft/sec2 = 9.8m/sec2

ax = acceleration in the x direction

ay = acceleration in the y direction

az = acceleration in the z direction

Weight = mass * ag

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Load Transfer

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

• Suspension Geometry

• Front: Location of instant centers (Side View Swing Arm)

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

• Suspension Springs

• Front: Coils, Air Springs, leafs or Torsion bars and 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

22

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.

23

Load Transfer Control Devices Springs

• Along with the dampers (shock absorbers), springs transfer the load of the sprung mass of the car to the road surface. During maneuvers, depending on instant center locations, the springs and dampers transfer some portion of the (m x a), mass x acceleration, forces to the ground.

• Spring Rate is force per unit displacement for a suspension spring alone .

• For coil springs this is measured axially along the centerline.

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

• For leaf springs it is measured at the axle seat.

• The spring rate may be linear (force increases proportionally with displacement) or nonlinear (increasing or decreasing rate with increasing displacement).

• Units are typically lb/in.

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Load Transfer Control Devices Anti-roll bars

• [Drawing 1] shows how an anti-roll bar (ARB) is twisted when the body rolls in a turn. This creates forces at the four points where the bar is attached to the vehicle. The forces are shown in [Drawing 2]. Forces A on the suspension increase [load] transfer to the outside tire. Forces B on the frame resist body roll. The effect is a reduction of body roll and an increase in [load] transfer at the end of the chassis which has the anti-roll bar. Because the total [load] transfer due to centripetal force is not changed, the opposite end of the chassis has reduced [load] transfer. [6]

Drawing 2 Drawing 1 Direction of Turn

A A

B B

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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 vehicle suspension to consider when analyzing suspension compliance.

Load Transfer Control Devices

26

Load Transfer Control Devices Frame/Chassis Deflection

• All of the calculations shown in this presentation are made under the assumption that the frame or chassis is completely rigid (both in torsion and bending). Of course any flexing within the frame/chassis will adversely effect the performance of the suspension which is attached to it.

Sprung and Unsprung Weight

100% Unsprung weight includes the mass of the tires, rims, brake rotors, brake calipers, knuckle assemblies, and ball joints which move in unison with the wheels.

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

The % unsprung weight of the shocks, springs and anti-roll bar ends would be a function of their motion ratio/2 with the remainder as % sprung weight.

The rest of the mass is on the vehicle side of the springs is suspended and is 100% sprung weight.

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Sprung and Unsprung Weight

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Springs Motion Ratio

The shocks, springs, struts and anti-roll bars are normally mounted at some angle from the suspension to the chassis.

Motion Ratio: If you were to move the wheel 1 inch and the spring were to deflect 0.75 inches then the motion ratio would be 0.75 in/in.

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

Springs Motion Ratio

30

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)

Wheel Rates

31

Wheel Rates are calculated by taking the square of the motion ratio times the spring rate. Squaring the ratio is because the ratio has two effects on the wheel rate. The ratio applies to both the force and distance traveled.

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

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

• Then the motion ratio is factored again to account for the lever-arm force differential.

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 B perpendicular to the line AP. If you want to move A 1 in vertically upward, B would only move (1in)(MR) = 0.75 in. Since K is 100 lb/in, and B has only moved 0.75 in, there's a force at B of 75 lb. If you balance the moments about P, you get 75(B)=X(A), and we know B = 0.75A, so you get 75(0.75A) = X(A). A's cancel and you get X=75(0.75)=56.25. Which is [100(MR)](MR) or 100(MR)2.

Wheel Rate (lb/in) = (Motion Ratio)2* (Spring Rate)

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

Wh

eel R

ate

(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 for both 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

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

Wh

eel R

ate

(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 during one 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

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 = F

req

uen

cy (c

ycle

s/m

in)

x = Static Deflection (in)

Ride Natural Frequency vs. Static Wheel Deflection

x

188


35

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.


36

Once the motion ratios has been established, the front and rear spring rates can be optimized for a “flat” ride at a particular speed.

37


Part II

Longitudinal Load Transfer

38

Center of Gravity

39

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.

Center of Gravity Location

40


WR

l R l F

L

Cg

WF

SCALE SCALE

Center of Gravity Horizontal Plane Location

41


100%

tot

LFRFfront

W

WWW

100%

tot

RRLRrear

W

WWW

100%

tot

LRLFleft

W

WWW 100%

tot

RRRFright

W

WWW

100%

tot

LRRFdiag

W

WWW

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 lb

W RF = 880 lb W LF = 880 lb W RR = 891 lb W LR = 891 lb

43


LW

WL

W

W

tot

r

tot

f

f

1

LW

WL

W

W

tot

f

tot

rr

1

44


983542

178298

3542

17601in3.49

f

983542

176098

3542

17821in7.48

r

Example: C3 Corvette Upgrade W tot = 3542 lb

W F = 1760 lb W R = 1782 lb L = 98.0 in

45

Center of Gravity Height Location

LWWLMHorizontal fRR 0:

coscoscoss in0

coscoss in0:

LWLWLWhWor

LWWLWhWMRaised

ffRcg

ffRcgR

46


The center of gravity height above the spindle centerline is:

hW

LW

W

LWaboveh

tot

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

47


The center of gravity height above the spindle centerline is:

12*3542

9604*62.22

98.6sin*3542

98.6cos*98*62.2211.5

98.6cos

98.6sin

98

1298.6tan

hCGtotal on level ground is: (where rt = tire radius)

89.1112*3542

9604*62.22in17

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

48


49


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.

50


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 WAxleFront R

FRF

L W =W W = L WAxleRear F

RFR

51


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 WAxleFront FRF 176098

7.483542

lb =W W = L WAxleRear RFR 178298

3.493542

Example: C3 Corvette W T = 3542 lb

W F = 1760 lb W R = 1782 lb L = 98.0 in lF = 49.3 in lR = 48.7 in

52

Longitudinal Load Transfer 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

aW

L

h a m =

L

h F = W cg

g

xT

cgcgaT **

53

Longitudinal Load Transfer 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

aT 2.28698

17*)2.32/15(*3542


L = 98.0 in hcg = 17 in ax = 15 ft/sec2

ag = 32.2 ft/sec2

54

L

h

a

a

L W =WW = 0 = M(B) cg

g

xRTF *


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

55


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


W F = 1760 lb W = 286.2 lb l R = 48.7 in


ag = 32.2 ft/sec2

A B

WF - W WR + W

l R l F

L

acceleration

force

Cg accel Fa WT

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

56


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 =WW = 0 = M(A) cg

g

xFTR *

57


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 R = 1782 lb W = 286.2 lb l F = 49.3 in


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

58

Longitudinal Acceleration Pitch The vehicle pitch angle , if no suspension forces oppose it (no

anti-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 right wheels. Therefore vertical displacements (Z) of the sprung (body) mass at the wheel locations are as shown.

ZR ZF

Cg Cg

JOU

NC

E

JOU

NC

E

REB

OU

ND

REB

OU

ND

(Jounce and Rebound are also known as Bump and Droop)

59

Longitudinal Acceleration Pitch

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

= WrWfdeg

FRWrWfrad *

2/2/2/2/

ZR ZF

Cg Cg

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

60


in = Z in = 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.

9826.230

2/2.286

35.314

2/2.286

180

+ =

+ =

+ =

deg

radrad

Example: C3 Corvette Upgrade 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

ZR ZF

Cg Cg


KW = MR2 * KS

61

Suspension Geometry

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 Geometry Side View Geometry

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.

63 FVSA

Swing Arm

SVSA

Swing Arm

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

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 and differential 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 b

66

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.

Side View Swing Arm

Cg

e

d

L

Fza

Fxa

Fx

Fz

IC

h

b

r

67

Anti- Geometry

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-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.

69

Anti-Squat Geometry

70

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

JOU

NC

E

REB

OU

ND

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

b

r WT

100% Anti-squat

Line

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

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).

L

h =

d

re =

btan

Cg

e

d

L

Fza

Fxa

Fx

Fz

IC

h

b

r WT

100% Anti-squat

Line

b

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

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

xxa a a

W = F = F

Anti-squat geometry

(1)

(2)

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

re F = F

0 = re F - d F0 = M

xaza

xazaR

d

re a

a

W = F x

g

T

za

Anti-squat geometry

(3)

(4)

Cg

e

d

L

Fza

Fxa

Fx

Fz

IC

h

b

r WT

100% Anti-squat

Line

b

77

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

re =

L

h

d

re a

a

W =

L

h a

a

W = F x

g

Tx

g

T

za

L

h =

d

re =

btan

Anti-squat geometry

(5)

(6)

78

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

L

h =

d

re =

btan

Anti-squat geometry

Cg

e

d

L

Fza

Fxa

Fx

Fz

IC

h

b

r WT

100% Anti-squat

Line

b

79

Anti-squat rear solid axle

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

Anti-squat geometry

Cg

e

d

L

Fza

Fxa

Fx

Fz

IC

h b

L

h

d

e = squatAnti %100

100*/

/

/

tan%

Lh

de

Lh = squatAnti

b

80

Anti-squat rear solid axle

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

Anti-squat geometry

Cg

e

d

L

Fza

Fxa

Fx

Fz

IC

h b

%125100*100/20

45/25.11

/

tan%

Lh = squatAnti

b

L

h

d

e = squatAnti %100

Example: Solid Axle e = 11.25 in

d = 45 in L = 100 in h = 20 in

81

Anti-squat independent rear suspension

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

Anti-squat geometry

L

h

d

re = squatAnti

%100

100*/

/

/

tan%

Lh

dre

Lh = squatAnti

b

Cg

e

d

L

Fza

Fxa

Fx

Fz

IC

h

b

r WT

100% Anti-squat

Line

b

82

Anti-squat independent rear suspension

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

Anti-squat geometry

L

h

d

re = squatAnti

%100

Example: C3 Upgrade e = 15.62 in

d = 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

b

Cg

e

d

L

Fza

Fxa

Fx

Fz

IC

h

b

r WT

100% Anti-squat

Line

b

83

Anti-squat geometry

Anti-squat effects on longitudinal pitch angle

100*%

*

=oncompensati pitch

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

84

Anti-squat geometry


Example: C3 Corvette Upgrade 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(

1180*

4.386

3542*

98

1

=(rad)anglePitch

squatAntidegrad =(deg)anglePitch %[email protected]

*0079.0

%28100*4103.

1136.% =oncompensati pitch

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

85

Anti-squat geometry


Example: C3 Corvette Upgrade 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(

10*

)2*35.314(

1

98

17*

)2*35.314(

1180*

4.386

3542*

98

1

=(rad)anglePitch

squatAntidegrad =(deg)anglePitch %[email protected]

*011.0

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

86


in = Z in = 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.

9826.230

2/2.286

35.314

2/2.286

180

+ =

+ =

+ =

deg

radrad

Example: C3 Corvette Upgrade 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 DW = 286.2 lb

L = 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

Anti-dive Geometry

88

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.

Cg

JOU

NC

E

REB

OU

ND

89

Anti-dive geometry 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).

Cg

L

h

WT

WT (ax/ag) = braking force

+ load - load

L

h a

a

W =

L

h a m = load x

g

T

90

Anti-dive geometry 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).

lb = L

h a m = load 2.286

98

17*180*

4.386

3542

Example: C3 Corvette Upgrade WT = 3542 lb

L = 98 in h = 17 in ax = 180 in/sec2 (.46 g)

ag = 386.4 in/sec2

91

Anti-dive geometry

Example: C3 Corvette Upgrade 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

Axl

e L

oad

(lb

f)

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

92

Anti-dive geometry

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

Cg

JOU

NC

E

REB

OU

ND

93

Anti-dive geometry

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.

Cg

JOU

NC

E

REB

OU

ND

94

Anti-dive geometry

Brake Bias (brake force distribution)

Cg

JOU

NC

E

REB

OU

ND

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

% o

f V

ehic

le L

oad

Tra

nsf

er

% o

f To

tal B

raki

ng

Forc

e

Deceleration (g's)

Typical Brake System Bias

% Front Load

% Front Braking

% Rear Load

% Rear Braking

95

Anti-dive geometry

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

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%.

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% brakingfronthL = liftAnti r b

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%.

96

Anti-dive geometry

Example: C3 Corvette Upgrade tan bf = .1124

tan br = .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 (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 rate of deceleration changes.

97

Anti-dive geometry

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

% A

nti

-div

e &

An

ti-l

ift

% o

f To

tal B

raki

ng

Forc

e

Deceleration (g's)

Anti-dive & Anti-lift vs. Brake Bias

% Anti-dive

% Front Braking

% Anti-lift

% Rear Braking

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.

99

Design factors in Anti-dive and Anti-Squat

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.

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 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).

101

Next… Part III - Lateral Load Transfer

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vehicle longitudinal load transfer parts i_ii_mar13

Automotive

vehicle dynamics

modern vehicle suspension

vehicle architecture

tires vehicle weight

rear suspension configuration

rear suspension system

roll bar factors

g factors