Fundamentals of Vehicle Dynamics

Fundamentals of Vehicle Dynamics
Fundamentals of Vehicle Dynamics
Topics Covered
INTRODUCTION
Dawn of the Motor Vehicle Age
Introduction to Vehicle Dynamics
Fundamental Approach to Modeling
Lumped Mass
Vehicle Fixed Coordinate System
Motion Variables
Earth Fixed Coordinate System
Euler Angles
Forces
Newton's Second Law
Dynamic Axle Loads
Static Loads on Level Ground
Low-Speed Acceleration
Loads on Grades
Example Problems
References
ACCELERATION PERFORMANCE
Power-Limited Acceleration
Engines
Power Train
Automatic Transmissions
Example Problems
Traction-Limited Acceleration
Transverse Weight Shift due to Drive Torque
Traction Limits
Example Problems
References
BRAKING PERFORMANCE
Basic Equations
Constant Deceleration
Deceleration with Wind Resistance
Energy/Power
Braking Forces
Rolling Resistance
Aerodynamic Drag
Driveline Drag
Grade
Brakes
Brake Factor
Tire-Road Friction
Velocity
Inflation Pressure
Vertical Load
Example Problems
Federal Requirements for Braking Performance
Brake Proportioning
Anti-Lock Brake Systems
Braking Efficiency
Rear Wheel Lockup
Pedal Force Gain
Example Problem
References
ROAD LOADS
Aerodynamics
Mechanics of Air Flow Around a Vehicle
Pressure Distribution on a Vehicle
Aerodynamic Forces
Drag Components
Aerodynamics Aids
·Bumper Spoilers
·Air Dams
·Deck Lid Spoilers
·Window and Pillar Treatments
·Optimization
Drag
·Air Density
·Drag Coefficient
Side Force
Lift Force
Pitching Moment
Yawing Moment
Rolling Moment
Crosswind Sensitivity
Rolling Resistance
Factors Affecting Rolling Resistance
·Tire Temperature
·Tire Inflation Pressure/Load
·Velocity
·Tire Material and Design
·Tire Slip
Typical Coefficients
Total Road Loads
Fuel Economy Effects
Example Problems
References
RIDE
Excitation Sources
Road Roughness
Tire/Wheel Assembly
Driveline Excitation
Engine/Transmission
Vehicle Response Properties
Suspension Isolation
Example Problem
Suspension Stiffness
Suspension Damping
Active Control
Wheel Hop Resonances
Suspension Nonlinearities
Rigid Body Bounce/Pitch Motions
Bounce/Pitch Frequencies
Special Cases
Example Problem
Perception of Ride
Tolerance to Seat Vibrations
Other Vibration Forms
Conclusion
References
STEADY STATE CORNERING
Introduction
Low-Speed Turning
High-Speed Cornering
Tire Cornering Forces
Cornering Equations
Understeer Gradient
Characteristic Speed
Critical Speed
Lateral Acceleration Gain
Yaw Velocity Gain
Sideslip Angle
Static Margin
Suspension Effects on Cornering
Roll Moment Distribution
Camber Change
Roll Steer
Lateral Force Compliance Steer
Aligning Torque
Effect of Tractive Forces on Cornering
Summary of Understeer Effects
Experimental Measurement of Understeer Gradient
Constant Radius Method
Constant Speed Method
Example Problems
References
SUSPENSIONS
Solid Axles
Hotchkiss
Four Link
De Dion
Independent Suspensions
Trailing Arm Suspension
SLA Front Suspension
MacPherson Strut
Multi-Link Rear Suspension
Trailing-Arm Rear Suspension
Semi-Trailing Arm
Swing Axle
Anti-Squat and Anti-Pitch Suspension Geometry
Equivalent Trailing Arm Analysis
Rear Solid Drive Axle
Independent Rear Drive
Front Solid Drive Axle
Independent Front-Drive Axle
Four-Wheel Drive
Anti-Dive Suspension Geometry
Example Problems
Roll Center Analysis
Solid Axle Roll Centers
·Four-Link Rear Suspension
·Three-Link Rear Suspension
·Four-Link with Parallel Arms
·Hotchkiss Suspension
Independent Suspension Roll Centers
·Positive Swing Arm Geometry
·Negative Swing Arm Geometry
·Parallel Horizontal Links
·Inclined Parallel Links
·MacPherson Strut
·Swing Axle
Active Suspensions
Suspension Categories
Functions
Performance
References
THE STEERING SYSTEM
Introduction
The Steering Linkages
Steering Geometry Error
Toe Change
Roll Steer
Front Wheel Geometry
Steering System Forces and Moments
Vertical Force
Lateral Force
Tractive Force
Aligning Torque
Rolling Resistance and Overturning Moments
Steering System Models
Examples of Steering System Effects
Steering Ratio
Understeer
Braking Stability
Influence of Front-Wheel Drive
Driveline Torque About the Steer Axis
Influence of Tractive Force on Tire Cornering Stiffness
Influence of Tractive Force on Aligning Moment
Fore/Aft Load Transfer
Summary of FWD Understeer Influences
Four-Wheel Steer
Low-Speed Turning
High-Speed Cornering
References
ROLLOVER
Quasi-Static Rollover of a Rigid Vehicle
Quasi-Static Rollover of a Suspended Vehicle
Transient Rollover
Simple Roll Models
Yaw-Roll Models
Tripping
Accident Experience
References
TIRES
Tire Construction
Size and Load Rating
Terminology and Axis System
Mechanics of Force Generation
Tractive Properties
Vertical Load
Inflation Pressure
Surface Friction
Speed
Relevance to Vehicle Performance
Cornering Properties
Slip Angle
Tire Type
Load
Inflation Pressure
Size and Width
Tread Design
Other Factors
Relevance to Vehicle Performance
Camber Thrust
Tire Type
Load
Inflation Pressure
Tread Design
Other Factors
Relevance to Vehicle Performance
Aligning Moment
Slip Angle
Path Curvature
Relevance to Vehicle Performance
Combined Braking and Cornering
Friction Circle
Variables
Relevance to Vehicle Performance
Conicity and Ply Steer
Relevance to Vehicle Performance
Durability Forces
Tire Vibrations
References
Acceleration Performance — Example from Chapter 2
Traction-Limited Acceleration
Problem
Find the traction-limited acceleration for a rear-drive passenger car with and without a locking differential based on the
following information:
Weight
Front
2100 lb
Rear
1850 lb
CG Height
21 in
Wheelbase
108 in
Coefficient of Friction
0.62
Tire size
13.0 in
Tread
59.0 in
Final drive ratio
2.90
Roll stiffness
Front
1150 lb-ft/deg
Rear
280 lb-ft/deg
Model Used
US CUSTOMARY UNITS >> CHAPTER 2 >> Traction Limited
Solution
When filled in ready to solve the Variable Sheet should look like the following one.
St Input
2100
1850
108
21
.62
13
59
2.9
1150
280
Name
Unit
Comment
Fundamentals of Vehicle Dynamics
by Thomas D. Gillespie
Chapter 2: Acceleration Performance
Traction Limits
Page 39, Eqs. 2-23,24,25,26
Wf
Wr
W
L
b
c
h
mu
lb
lb
lb
in
in
in
in
Weight on front axle
Weight on rear axle
Total weight
Wheelbase
CG to front axle
Rear axle to CG
CG height
Coefficient of friction
r
t
Nf
Kphif
Kphir
Kphi
in
in
lb-ft/deg
lb-ft/deg
lb-ft/deg
Fxmax1
axmax1
lb
g's
Fxmax2
axmax2
lb
g's
Fxmax3
axmax3
lb
g's
Fxmax4
axmax4
lb
g's
Click on
St Input
2100
1850
Non-locking differential parameters:
Radius of tires
Tread
Final drive ratio
Front suspension roll stiffness
Rear suspension roll stiffness
Total roll stiffness
Solid non-locking rear axle:
Maximum tractive force
Maximum acceleration
Solid locking/independent rear axle:
Maximum tractive force
Maximum acceleration
Solid non-locking front axle:
Maximum tractive force
Maximum acceleration
Solid locking/independent front axle:
Maximum tractive force
Maximum acceleration
or press F9 to solve the model. The Variable Sheet should then appear as follows.
Name
21
.62
Wf
Wr
W
L
b
c
h
mu
13
59
r
t
108
Output
Output
3950
50.582278
57.417722
Unit
Comment
Fundamentals of Vehicle Dynamics
by Thomas D. Gillespie
Chapter 2: Acceleration Performance
Traction Limits
Page 39, Eqs. 2-23,24,25,26
lb
lb
lb
in
in
in
in
Weight on front axle
Weight on rear axle
Total weight
Wheelbase
CG to front axle
Rear axle to CG
CG height
Coefficient of friction
in
in
Non-locking differential parameters:
Radius of tires
Tread
2.9
1150
280
Nf
Kphif
Kphir
Kphi
1430
lb-ft/deg
lb-ft/deg
lb-ft/deg
Fxmax1
axmax1
1200.782
.30399544
lb
g's
Fxmax2
axmax2
1304.2325
.33018544
lb
g's
Fxmax3
axmax3
1143.1049
.28939366
lb
g's
Fxmax4
axmax4
1161.9236
.29415789
lb
g's
Final drive ratio
Front suspension roll stiffness
Rear suspension roll stiffness
Total roll stiffness
Solid non-locking rear axle:
Maximum tractive force
Maximum acceleration
Solid locking/independent rear axle:
Maximum tractive force
Maximum acceleration
Solid non-locking front axle:
Maximum tractive force
Maximum acceleration
Solid locking/independent front axle:
Maximum tractive force
Maximum acceleration
Braking Performance — Example from Chapter 3
Braking Coefficients and Efficiency
Problem
Calculate the braking coefficients and braking efficiency for a passenger car in 100 psi increments of application pressure
up to 700 psi given the following information:
Wheelbase
108.5 inches
Center of gravity height
20.5 inches
Tire radius
12.11 inches
Weights
Front
2210 lb
Rear
1864 lb
Total
4074 lb
Brake Gain
Front
20 in-lb/psi
Rear
14 in-lb/psi
Proportioning change pressure
290
Rate adjustment factor
.03
Model Used
US CUSTOMARY UNITS >> CHAPTER 3 >> Efficiency and Braking Coefficient Plot
Solution
This model uses several of TK Solver’s features. The Variable Sheet holds the constants for the problem. A table holds
the values of application pressure (Pa) being considered as well as the associated values calculated for rear application
pressure (Pr), front and rear brake forces (Ff, Fr), deceleration (Dx), front and rear axle loads (Wf, Wr), braking coefficients
( f, r), and braking efficiency ( b). There are also four plots available. The first shows the deceleration and efficiency
over the range of application pressure as per Figure 3.12 of the text. The second shows the front and rear braking
coefficients over the range of ap plication pressures as per Figure 3.13 of the text. The third and fourth show the efficiency
and the braking coefficients vs. deceleration.
The constants from the table shown above are entered on the Variable Sheet. Note that the model does not require that
you enter all three values for front, rear, and total axle static load but only two of the three. In this example, values were
entered for front and rear axle static load. When all values have been entered, the Variable Sheet should look like the
following one. (The outputs will appear after invoking the Solve command as described later on.)
St Input
Name
Output
Unit
Comment
Fundamentals of Vehicle Dynamics
by Thomas D. Gillespie
Chapter 3: Braking Performance
Fig. 3.12 Braking efficiency vs. Pa
Fig. 3.13 Braking coefficient vs. Pa
Braking efficiency vs. deceleration
Braking coefficient vs. deceleration
Enter application pressures in table,
other values on Variable Sheet.
108.5
20.5
12.11
2210
1864
20
14
290
.3
L
h
r
Wfs
Wrs
W
Gf
Gr
Pdelta
raf
g
4074
1
in
in
in
lb
lb
lb
in-lb/psi
in-lb/psi
psi
g's
Wheelbase
CG height
Tire radius
Front axle static load
Rear axle static load
Total load
Front brake gain (per brake)
Rear brake gain (per brake)
Proportioning change pressure
Rate adjustment factor
Acceleration of gravity: default = 1 g
The application pressures are entered in the model’s table. The table should already be open and only need a mouse
click to gain focus.
Opening an Interactive Table
If a table has been minimized, there will be an icon at the bottom of the TK Solver window that looks similar to the
following one.
Double-clicking on the icon will restore the table window.
If a table is neither open nor minimized, or if its icon is hidden by open windows, you can open it by selecting it from the
Tables listbox in the Object bar.
The values for the application pressure are entered in the Pa column of the table. The values can be entered manually
one at a time or automatically using TK’s List Fill command. The following steps would allow you to fill the list with TK’s
List Fill command.
1. Click on
in the Toolbar, or choose the List Fill option from the Commands menu.
2. Enter Pa in the List Name field. You can also select the name through a listbox activated by pressing the button at
the right end of the field.
3. The List Fill dialog gives you many ways to place values into lists. For this example fill in the dialog as shown
below. Press the Fill List button when finished.
Once the constants and application pressures have been input, click on
model. The table should be filled in as shown below.
in the Toolbar or press F9 to solve the
Pa psi
100
200
300
400
500
600
700
mur
.132
.280
.437
.507
.584
.670
.766
Pr psi
100
200
293
323
353
383
413
Fxf lb
330
661
991
1321
1652
1982
2312
Fxr lb
231
462
677
747
816
886
955
Dx g's
.138
.276
.410
.508
.606
.704
.802
Wf lb
2316
2422
2525
2601
2676
2752
2827
Wr lb
1758
1652
1549
1473
1398
1322
1247
muf
.143
.273
.392
.508
.617
.720
.818
etab %
96.6
98.5
93.6
99.9
98.2
97.7
98.1
This model treats the Pa column as input and all other columns as output. You can make changes to the values in the
other columns, but the next time you solve the model they will all be overwritten by values determined by the entries in the
Pa column.
Closing an Interactive Table
To close a window on an interactive table, click on the Minimize Window-Sizing Icon (the down-arrow button in the upper
right corner of the window). This will minimize the table as an icon that you can double-click to reopen.
You can also single-click the System Menu Icon (the button in the upper left corner of the table window) and select
Minimize from the resulting menu. If you double-click the System Menu Icon the table window will close all right but you
will also lose the table icon.
If you inadvertently double-click the System Menu Icon to close the table window, or if the table icon is hidden by open
windows, follow the steps outlined in Opening an Interactive Table to reopen the table.
The plots for this example are shown below.
Closing a Plot
To close a window on a plot, click on the Minimize Window-Sizing Icon (the down-arrow button in the upper right corner of
the window). This will minimize the plot as an icon that you can double-click to redisplay.
You can also single-click the System Menu Icon (the button in the upper left corner of the plot window) and select
Minimize from the resulting menu. If you double-click the System Menu Icon the plot window will close all right but you will
also lose the plot icon.
If you inadvertently double-click the System Menu Icon to close the plot window, or if the plot icon is hidden by open
windows, follow the steps outlined in Displaying a Plot to redisplay the plot.
Road Loads — Example from Chapter 4
Forces and Moments
Problem
A passenger car has a fontal area of 21 square feet and a drag coefficient of 0.42. It is traveling at 55 mph. Calculate the
aerodynamic drag for the cases of a 25 mph headwind and a 25 mph tailwind.
Model Used
US CUSTOMARY UNITS >> CHAPTER 4 >> Forces and Moments
Solution
Since TK Solver does not require a specific set of inputs we can use this model to gain the information we need even
though it contains variables and equations that are not related to the problem at hand. During the solution process, TK will
solve for as many unknowns as possible. This problem requires only three inputs on the Variable Sheet as shown below.
For the case of the headwind the expression 55 + 25 can be entered as input for the variable V.
St Input
Name
Output
Unit
Comment
Fundamentals of Vehicle Dynamics
by Thomas D. Gillespie
Chapter 4: Road loads
Aerodynamics
Eqs. 4-2, 4-6, 4-7, 4-8, 4-9, 4-10
21
80
.42
L
A
V
Tr
Pr
rho
.002378
in
ft^2
mph
F
in Hg
slug/ft^3
q
16.369084
lb/ft^2
Coefficients:
Aerodynamic drag
Side force
Lift
Pitching moment
Yawing moment
Rolling moment
CD
CS
CL
CPM
CYM
CRM
DA
SA
LA
Wheelbase
Frontal area of vehicle
Total wind velocity
Air temperature
Atmospheric pressure
Air density (if Tr or Pr not given,
default = .002378 slug/ft^3)
Dynamic pressure, 1/2*rho*V^2
144.37532
lb
lb
lb
Drag force
Side force
Lift force
PM
YM
RM
lb-ft
lb-ft
lb-ft
Pitching moment
Yawing moment
Rolling moment
The aerodynamic drag, DA, for the headwind situation is approximately 144 pounds. To find the solution for the tailwind
situation enter the expression 55 – 25 for the variable V and solve the model again. The aerodynamic drag will be
approximately 20 pounds. Note that you are not stuck with the default air density. You can enter a different value in the
Input field for rho. You can also enter inputs for any two of Tr, Pr and rho, and TK will calculate the value of the third one.
Ride — Example from Chapter 5
Pitch and Bounce Centers and Frequencies
Problem
Calculate the pitch and bounce centers and frequencies for a car with the following characteristics.
Ride rate
Front
127 lb/in
Rear
92.3 lb/in
Tire load
Front
957 lb
Rear
730 lb
Wheelbase
100.6
Dynamic Index
1.1
Model Used
US CUSTOMARY UNITS >> CHAPTER 5 >> Bounce/Pitch Frequencies
Solution
Fill in the known values as inputs on the model’s Variable Sheet, then solve the model (
or F9). The Variable Sheet
will look like the following one after solving.
St Input
Name
Output
Unit
Comment
Fundamentals of Vehicle Dynamics
by Thomas D. Gillespie
Chapter 5: Ride
Vehicle Response Properties
Bounce/Pitch Frequencies
Fig. 5.34 Natural Frequency Ratio
vs. Motion Centers
127
92.3
957
730
ff%frmi
ff%frma
.8
1.2
g
386.08858
nf
nr
Kf%t
Kr%t
Kf
Kr
Wf%t
Wr%t
Wf
Wr
W
M
2
2
254
184.6
1914
1460
3374
8.7389272
Plot parameters:
Minimum ff/fr (default = .8)
Maximum ff/fr (default = 1.2)
in/s^2
lb/in
lb/in
lb/in
lb/in
lb
lb
lb
lb
lb
lb-s^2/in
Acceleration due to gravity
(default = 386.08858 in/s^2)
Number of front tires (default = 2)
Number of rear tires (default = 2)
Front ride rate per tire
Rear ride rate per tire
Total front ride rate
Total rear ride rate
Front tire load per tire
Rear tire load per tire
Total front tire load
Total rear tire load
Weight of vehicle
Mass of vehicle
100.6
1.1
L
b
c
DI
k
43.531713
57.068287
in
in
in
Wheelbase
Front axle to CG
CG to rear axle
Dynamic Index
Radius of gyration (k^2 = b*c*DI)
52.275313
in
Z%t1
Z%t2
omega1
omega2
ff
fr
ff%fr
-233.8588
11.685289
1.1303897
1.0685331
1.139225
1.1119955
1.024487
in/rad
in/rad
Hz
Hz
1/s
1/s
Centers and Frequencies:
CG to oscillation center 1
CG to oscillation center 2
Frequency 1
Frequency 2
Front suspension natural frequency
Rear suspension natural frequency
Ratio ff/fr
k~2
alpha
beta
gamma
Iy
2732.7083
50.189227
-59.76127
45.33054
23880.939
in^2
1/s^2
in/s^2
1/s^2
in-lb-s^2
Intermediate Variables:
k^2
(2*Kf+2*Kr)/M
(2*Kr*c-2*Kf*b)/M
(2*Kf*b^2+2*Kr*c^2)/(M*k^2)
Pitch moment of inertia
(Iy=M*k^2)
The model also has a plot available. If it is minimized at the bottom of the TK Window its icon will look like the following.
You can open the plot by double-clicking the icon or pressing the F7 special function key.
Because there is only one plot in this model, you don’t need to select it from the Plots listbox on the Object bar. You only
need to press F7 if you cannot see the plot icon. The plot will look like the following one.
TK consistently plots negative numbers on the left and positive numbers on the right. The vertical lines marking the wheel
locations are labeled with f and r to indicate the orientation of the front and rear wheels.
Steady-State Cornering — Example from Chapter 6
Steer Angle and Yaw Velocity Gain vs. Speed
Problem
A vehicle has a wheelbase of 104 in and a radius of turn of 100 feet. Analyze the change in steer angle and yaw velocity
gain from 0 to 125 mph under the following conditions.
Dynamic Weight on Axle (lb)
Front
Rear
Case 1 776
950
Case 2 950
776
Case 3 800
800
Cornering Stiffness of Tires (lb/deg)
Front
Rear
300
275
100
95
150
150
Model Used
US CUSTOMARY UNITS >> CHAPTER 6 >> Turning Response - High Speed
Solution
The Variable Sheet holds the minimum and maximum speeds for the plot range and the acceleration due to gravity. They
are provided with defaults as shown in the following Variable Sheet, so you don’t have to enter any values on the Variable
Sheet unless you want to change the defaults.
St Input
Name
Output
Unit
Comment
Fundamentals of Vehicle Dynamics
by Thomas D. Gillespie
Chapter 6: Steady-State Cornering
High-Speed Cornering
Fig. 6.5 Steer angle vs. speed
Fig. 6.6 Yaw velocity gain vs. speed
Sideslip angle vs. speed
Enter up to 5 cases in Input table
Vmin
Vmax
0
125
mph
mph
g
1
g's
Plot parameters:
Minimum speed (default = 0 mph)
Maximum speed (default = 125 mph)
Acceleration due to gravity
(default = 1 g)
The values for the radius of turn, wheelbase, dynamic weights on the front and rear axles, and cornering stiffnesses of the
tires are entered into an interactive table named Inputs. While this example uses three cases, the model will plot up to ten
different cases at one time. If the table isn’t already visible, it may be minimized at the bottom of the TK Solver window. If
so, the icon will look like the following and double-clicking on the icon will open the table.
If neither the table nor its icon is visible, select the table from the Tables listbox on the Object bar.
Fill in the values for the different cases. When complete the Input table will look like the following.
R ft
100.00
100.00
100.00
L in
104.00
104.00
104.00
Wf lb
776.00
950.00
800.00
Wr lb
950.00
776.00
800.00
Caf lb/deg
300.00
100.00
150.00
Car lb/deg
275.00
95.00
150.00
Once the values for the problems have been entered into the Variable Sheet and the table, invoke the Solve command
from the Commands menu. You may also start the Solve command by clicking on the single light bulb button
Toolbar or by pressing the F9 special function key.
on the
The model will automatically generate two plots as per figures 6.5 and 6.6 in the text. It will also generate a plot for beta
vs. speed not shown in the text. If the plots are minimized at the bottom of the window, they will have the following icons
and can be viewed by double-clicking on the icon. (You can also display a plot by clicking on the plot button
in the
toolbar or by pressing function key F7. The plot displayed depends on which icon is currently highlighted or, if no icon is
highlighted, the row that was highlighted the last time the Plot Sheet was the active window.)
If the plot icons are not visible, follow the steps in Displaying a Plot. The plots for this example are shown below.
The characteristic speeds for understeer cases and critical speeds for oversteer cases are determined in Fig. 6.5 by the
points at which the lines end at the top and bottom of the plot, respectively.
In Fig. 6.6 the characteristic speed for an understeer case is indicated by a vertical line dropping down from the peak of
the curve. The critical speed for an oversteer case is indicated by a vertical line through the entire height of the plot. If
there are multiple cases of oversteer, the quickest way to match critical speeds with the curves is by color. If color is not
available, the steepest curve is matched with the leftmost vertical line. (The vertical lines associated with shallower curves
may be beyond the speed range shown on the plot.)
Suspensions — Example from Chapter 7
Anti-Squat & Anti-Pitch Suspension Geometry, Rear-Wheel, Solid Axle Drive.
Problem
Find the geometry that would be necessary to achieve 100% anti-squat in the rear suspension, and the geometry to
achieve full anti-pitch for the solid-axle, rear-wheel-drive vehicle described below. Also, find the pitch rate (degrees pitch/g
acceleration) when the geometry is set for 100% anti-squat in the rear suspension.
Suspension Spring Rates
Front
285 lb/in
Rear
169 lb/in
Center of Gravity Height 20.5 in
Wheelbase
108.5 in
Weight
4074 lb
Model Used
US CUSTOMARY UNITS >> CHAPTER 7 >> Rear Solid Drive Axle
Solution
The variable e%d represents the ratio e/d, and the comment for e%d indicates that the default will be
h/L + h/L*Kr/Kf
if either or both of e and d are unknown. This expression is the right hand side of Equation 7-15, so you can find the
solution to the second case (full anti-pitch) by not giving inputs to both e and d. If you give one of them an input, the other
will be calculated as an output dependent on the default for e%d. To explore this feature, give e an input value of 10.
Since full anti-pitch means that thetap will be extremely close to zero no matter what acceleration the vehicle is
undergoing, the input for ax can be any arbitrary value. Setting ax = 1 g will make it easy to translate pitch angle thetap
into a pitch rate for the 100% anti-squat case to be solved later.
When you have finished entering data and solving, your Variable Sheet should look like the following one.
St Input
1
4074
108.5
20.5
169
285
10
Name
Output
Unit
Comment
Fundamentals of Vehicle Dynamics
by Thomas D. Gillespie
Chapter 7: Suspensions
Anti-Squat and Anti-Pitch Suspension
Geometry
Rear Solid Drive Axle
Page 251, Eqs. 7-14, 7-15
g
ax
W
L
h
Kr
Kf
e
d
e%d
1
33.224992
.30097825
g's
g's
lb
in
in
lb/in
lb/in
in
in
thetap
9.574E-16
Acceleration of gravity: default = 1 g
Acceleration in x-direction
Total weight
Wheelbase
CG height
Total rear suspension spring rate
Total front suspension spring rate
Height of rear imaginary pivot
Imaginary pivot to rear wheel
Ratio e/d (if e and/or d unknown,
default = h/L + h/L*Kr/Kf)
Pitch angle of vehicle
deg
Full anti-pitch therefore occurs when the ratio e/d is approximately .301. The 100% anti-squat case occurs when
e/d = h/L
This condition can be met simply by typing h/L in the Input field for the variable e%d. When you press Enter (or click the
mouse pointer on another field, or use one of the navigation arrow keys), TK evaluates the expression and enters the
result in the field.
When you have finished giving an input to e%d and solving, the Variable Sheet for the 100% anti-squat case should look
like the following one.
St Input
1
4074
108.5
20.5
169
285
10
.18894009
Name
Output
Unit
Comment
Fundamentals of Vehicle Dynamics
by Thomas D. Gillespie
Chapter 7: Suspensions
Anti-Squat and Anti-Pitch Suspension
Geometry
Rear Solid Drive Axle
Page 251, Eqs. 7-14, 7-15
g
ax
W
L
h
Kr
Kf
e
d
e%d
1
52.926829
g's
g's
lb
in
in
lb/in
lb/in
in
in
1.4262419
deg
Acceleration of gravity: default = 1 g
Acceleration in x-direction
Total weight
Wheelbase
CG height
Total rear suspension spring rate
Total front suspension spring rate
Height of rear imaginary pivot
Imaginary pivot to rear wheel
Ratio e/d (if e and/or d unknown,
default = h/L + h/L*Kr/Kf)
Pitch angle of vehicle
thetap
Since ax is 1 g, the pitch rate is (1.4262419 deg)/(1g), or approximately 1.43 deg/g.
The Steering System — Example from Chapter 8
Steering Torque Arising From Lateral Inclination and Caster Angles
Problem
Reproduce Figures 8.12 and 8.14 of the text using the following data.
Vertical Load
Left
800 pounds
Right
600 pounds
Lateral Inclination Angle
10 degrees
Lateral Offset
1 inch
Caster Angle
5 degrees
Minimum Steer Angle
-45 degrees
Maximum Steer Angle
45 degrees
Model Used
US CUSTOMARY UNITS >> CHAPTER 8 >> Steering Torque Due to Vertical Forces
Solution
All information in this model is entered on the Variable Sheet. Once the values for the problem have been entered into the
Variable Sheet, select the Solve command from the Commands menu. You may also start the Solve command by clicking
on the single light bulb button
on the Toolbar or by pressing function key F9.
When you have finished entering the data and solving the model, the Variable Sheet should look like the following.
St Input
Name
Output
Unit
Comment
Fundamentals of Vehicle Dynamics
by Thomas D. Gillespie
Chapter 8: The Steering System
Steering System Forces and Moments
Vertical Force
Fig. 8.12 Steering torque from
lateral inclination angle
Fig. 8.14 Steering torque from
caster angle
mindelta
maxdelta
800
600
1
10
-45
5
-45
45
Fzl
Fzr
d
lambda
delta
nu
MvLatL
MvLatR
MvLat
MvCasL
MvCasR
MvCas
Mv
98.230243
73.672682
171.90293
49.302733
-36.97705
12.325683
184.22861
deg
deg
Plot parameters:
Minimum steer angle (default = -45 deg)
Maximum steer angle (default = +45 deg)
lb
lb
in
deg
deg
deg
Vertical load on left wheel
Vertical load on right wheel
Lateral offset at ground
Lateral inclination angle
Steer angle
Caster angle
lb-in
lb-in
lb-in
lb-in
lb-in
lb-in
lb-in
Moments due to vertical force:
from lateral inclination, left wheel
from lateral inclination, right wheel
from lateral inclination, total
from caster, left wheel
from caster, right wheel
from caster, total
total moment from both wheels
The model will automatically generate two plots as per Figures 8.12 and 8.14 in the text. If the plots are minimized at the
bottom of the window, they will have the following icons and can be viewed by double-clicking on the icon.
If the plot icons are not visible, follow the steps in Displaying a Plot. The plots for this example are shown below.
Rollover — Example from Chapter 9
Road Cross Slope
Problem
Given the following characteristics of a vehicle in a turn, find the cross-slope angle phi such that the occupants experience
a lateral acceleration of .1 g, then backsolve for the neutral speed for that cross-slope.
Velocity
40 mph
Radius of Turn
500 feet
Track width (tread)
60 inches
Height of Center of Gravity
20 inches
Weight
2200 pounds
Model Used
US CUSTOMARY UNITS >> CHAPTER 9 >> Rigid Vehicle
Solution
This model makes a distinction between ac, the centrifugal acceleration due to the vehicle traveling a curved path, and ay,
the lateral acceleration experienced by the occupants where "lateral" is relative to the vehicle coordinate axis system. The
road cross-slope angle, phi, has no affect on ac but has a strong affect on ay.
All inputs required for the solution are handled on the Variable Sheet. When you have finished entering the data, solve the
model by clicking on the single light bulb button
should then look like the following.
St Input
Name
Output
Unit
on the Toolbar or by pressing function key F9. The Variable Sheet
Comment
Fundamentals of Vehicle Dynamics
by Thomas D. Gillespie
Chapter 9: Rollover
Quasi-Static Rollover of Rigid Vehicle
Eq. 9-1, p. 311, and Fig. 9.2, p. 313
40
500
60
20
2200
.1
V
R
ac
t
h
W
g
M
1
2200
mph
ft
g's
in
in
lb
g's
lbm
Velocity
Radius of turn
Centrifugal (horizontal) acceleration
Track width (tread)
Height of CG
Weight of vehicle = M*g
Acceleration of gravity (default: 1 g)
Mass of vehicle
ay
az
Fzi
Fzo
phi
1.0177296
1046.1692
1192.8359
6.4644744
g's
g's
lb
lb
deg
Acceleration along vehicle y-axis
Acceleration along vehicle z-axis
Normal force on inside tire
Normal force on outside tire
Road cross-slope angle
Vroll
acroll
ayroll
azroll
120.5635
1.9436535
1.8187084
1.2124722
mph
g's
g's
g's
Rollover thresholds:
Velocity
Centrifugal acceleration
Acceleration along vehicle y-axis
Acceleration along vehicle z-axis
.21394745
The cross-slope angle is approximately 6.5 deg.
This model will automatically generate three plots. The first plot shows the equilibrium lateral acceleration in rollover of a
rigid vehicle on a cross-slope of 0 deg as per Figure 9.2 in the text. The second plot is similar except that the cross-slope
is taken as the angle phi given or calculated on the Variable Sheet. The third plot is a vector diagram showing the
relationship between the vectors ac and g (relative to horizontal and vertical) and ay and az (relative to the vehicle axis
system.) A plot can be displayed by double-clicking on its icon.
If a plot icon is not visible for some reason, follow the steps in Displaying a Plot to redisplay the plot. The plots for this
example are shown below.
The point common to the upper left corners of both rectangles represents the location of the center of gravity. The
horizontal line from the CG to the point labeled G represents the centrifugal acceleration, ay, while the vertical line from
the CG to the other point labeled G represents the acceleration due to gravity, g. On the other rectangle, the line slanting
right and slightly upwards from the CG to the point labeled V represents the lateral acceleration experienced by the
occupants, ay, while the line slanting down and slightly right from the CG to the other point labeled V represents the
vertical acceleration experienced by the occupants, az.
The neutral speed is the speed at which the occupants experience no lateral acceleration relative to the vehicle. To
backsolve for this case, change phi from an output variable to an input variable (quickest way: type I in the Status field),
change the input for ay from .1 to 0, and blank the Input field for V (type B in the Status field, or type Spacebar Enter over
the Input field). After making the changes and solving again, the Variable Sheet should look as follows.
St Input
Name
Output
Unit
Comment
Fundamentals of Vehicle Dynamics
by Thomas D. Gillespie
Chapter 9: Rollover
Quasi-Static Rollover of Rigid Vehicle
Eq. 9-1, p. 311, and Fig. 9.2, p. 313
500
60
20
2200
0
6.4644744
V
R
ac
t
h
W
g
M
29.109577
.11330756
1
2200
mph
ft
g's
in
in
lb
g's
lbm
Velocity
Radius of turn
Centrifugal (horizontal) acceleration
Track width (tread)
Height of CG
Weight of vehicle = M*g
Acceleration of gravity (default: 1 g)
Mass of vehicle
ay
az
Fzi
Fzo
phi
1.0063988
1107.0387
1107.0387
g's
g's
lb
lb
deg
Acceleration along vehicle y-axis
Acceleration along vehicle z-axis
Normal force on inside tire
Normal force on outside tire
Road cross-slope angle
Vroll
acroll
ayroll
azroll
120.5635
1.9436535
1.8187084
1.2124722
mph
g's
g's
g's
Rollover thresholds:
Velocity
Centrifugal acceleration
Acceleration along vehicle y-axis
Acceleration along vehicle z-axis
The neutral speed for this curve is approximately 29 mph. The equilibrium lateral acceleration plots will be the same for
this solution because the radius, cross-slope, track width (tread), and CG height are unchanged. The plot of acceleration
vectors is different because the centrifugal acceleration, ac, is smaller and the lateral acceleration ay is now 0.
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