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 Name Output Unit Comment Fundamentals of Vehicle Dynamics by Thomas D. Gillespie Chapter 2: Acceleration Performance Traction Limits Page 39, Eqs. 2-23,24,25,26 2100 Wf lb Weight on front axle 1850 Wr lb Weight on rear axle W lb Total weight 108 L in Wheelbase b in CG to front axle c in Rear axle to CG 21 h in CG height .62 mu Coefficient of friction Non-locking differential parameters: 13 r in Radius of tires 59 t in Tread 2.9 Nf Final drive ratio 1150 Kphif lb-ft/deg Front suspension roll stiffness 280 Kphir lb-ft/deg Rear suspension roll stiffness Kphi lb-ft/deg Total roll stiffness Solid non-locking rear axle: Fxmax1 lb Maximum tractive force axmax1 g's Maximum acceleration Solid locking/independent rear axle: Fxmax2 lb Maximum tractive force axmax2 g's Maximum acceleration Solid non-locking front axle: Fxmax3 lb Maximum tractive force axmax3 g's Maximum acceleration Solid locking/independent front axle: Fxmax4 lb Maximum tractive force axmax4 g's Maximum acceleration

Click on or press F9 to solve the model. The Variable Sheet should then appear as follows.

St Input Name Output Unit Comment Fundamentals of Vehicle Dynamics by Thomas D. Gillespie Chapter 2: Acceleration Performance Traction Limits Page 39, Eqs. 2-23,24,25,26 2100 Wf lb Weight on front axle 1850 Wr lb Weight on rear axle W 3950 lb Total weight 108 L in Wheelbase b 50.582278 in CG to front axle c 57.417722 in Rear axle to CG 21 h in CG height .62 mu Coefficient of friction Non-locking differential parameters: 13 r in Radius of tires 59 t in Tread

2.9 Nf Final drive ratio 1150 Kphif lb-ft/deg Front suspension roll stiffness 280 Kphir lb-ft/deg Rear suspension roll stiffness Kphi 1430 lb-ft/deg Total roll stiffness Solid non-locking rear axle: Fxmax1 1200.782 lb Maximum tractive force axmax1 .30399544 g's Maximum acceleration Solid locking/independent rear axle: Fxmax2 1304.2325 lb Maximum tractive force axmax2 .33018544 g's Maximum acceleration Solid non-locking front axle: Fxmax3 1143.1049 lb Maximum tractive force axmax3 .28939366 g's Maximum acceleration Solid locking/independent front axle: Fxmax4 1161.9236 lb Maximum tractive force axmax4 .29415789 g's 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 L in Wheelbase 20.5 h in CG height 12.11 r in Tire radius 2210 Wfs lb Front axle static load 1864 Wrs lb Rear axle static load W 4074 lb Total load 20 Gf in-lb/psi Front brake gain (per brake) 14 Gr in-lb/psi Rear brake gain (per brake) 290 Pdelta psi Proportioning change pressure .3 raf Rate adjustment factor g 1 g's 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 in the Toolbar or press F9 to solve the model. The table should be filled in as shown below.

Pa psi Pr psi Fxf lb Fxr lb Dx g's Wf lb Wr lb muf mur etab % 100 100 330 231 .138 2316 1758 .143 .132 96.6 200 200 661 462 .276 2422 1652 .273 .280 98.5 300 293 991 677 .410 2525 1549 .392 .437 93.6 400 323 1321 747 .508 2601 1473 .508 .507 99.9 500 353 1652 816 .606 2676 1398 .617 .584 98.2 600 383 1982 886 .704 2752 1322 .720 .670 97.7 700 413 2312 955 .802 2827 1247 .818 .766 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 L in Wheelbase 21 A ft^2 Frontal area of vehicle 80 V mph Total wind velocity Tr F Air temperature Pr in Hg Atmospheric pressure rho .002378 slug/ft^3 Air density (if Tr or Pr not given, default = .002378 slug/ft^3) q 16.369084 lb/ft^2 Dynamic pressure, 1/2*rho*V^2 Coefficients: .42 CD Aerodynamic drag CS Side force CL Lift CPM Pitching moment CYM Yawing moment CRM Rolling moment DA 144.37532 lb Drag force SA lb Side force LA lb Lift force

PM lb-ft Pitching moment YM lb-ft Yawing moment RM lb-ft 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 Plot parameters: ff%frmi .8 Minimum ff/fr (default = .8) ff%frma 1.2 Maximum ff/fr (default = 1.2) g 386.08858 in/s^2 Acceleration due to gravity (default = 386.08858 in/s^2) nf 2 Number of front tires (default = 2) nr 2 Number of rear tires (default = 2) 127 Kf%t lb/in Front ride rate per tire 92.3 Kr%t lb/in Rear ride rate per tire Kf 254 lb/in Total front ride rate Kr 184.6 lb/in Total rear ride rate 957 Wf%t lb Front tire load per tire 730 Wr%t lb Rear tire load per tire Wf 1914 lb Total front tire load Wr 1460 lb Total rear tire load W 3374 lb Weight of vehicle M 8.7389272 lb-s^2/in Mass of vehicle

100.6 L in Wheelbase b 43.531713 in Front axle to CG c 57.068287 in CG to rear axle 1.1 DI Dynamic Index k 52.275313 in Radius of gyration (k^2 = b*c*DI) Centers and Frequencies: Z%t1 -233.8588 in/rad CG to oscillation center 1 Z%t2 11.685289 in/rad CG to oscillation center 2 omega1 1.1303897 Hz Frequency 1 omega2 1.0685331 Hz Frequency 2 ff 1.139225 1/s Front suspension natural frequency fr 1.1119955 1/s Rear suspension natural frequency ff%fr 1.024487 Ratio ff/fr Intermediate Variables: k~2 2732.7083 in^2 k^2 alpha 50.189227 1/s^2 (2*Kf+2*Kr)/M beta -59.76127 in/s^2 (2*Kr*c-2*Kf*b)/M gamma 45.33054 1/s^2 (2*Kf*b^2+2*Kr*c^2)/(M*k^2) Iy 23880.939 in-lb-s^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) Cornering Stiffness of Tires (lb/deg) Front Rear Front Rear

Case 1 776 950 300 275 Case 2 950 776 100 95 Case 3 800 800 150 150 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 Plot parameters: Vmin 0 mph Minimum speed (default = 0 mph) Vmax 125 mph Maximum speed (default = 125 mph) g 1 g's 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 L in Wf lb Wr lb Caf lb/deg Car lb/deg 100.00 104.00 776.00 950.00 300.00 275.00 100.00 104.00 950.00 776.00 100.00 95.00 100.00 104.00 800.00 800.00 150.00 150.00

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 on the Toolbar or by pressing the F9 special function key.

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/inCenter of Gravity Height 20.5 in Wheelbase 108.5 inWeight 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 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 1 g's Acceleration of gravity: default = 1 g 1 ax g's Acceleration in x-direction 4074 W lb Total weight 108.5 L in Wheelbase 20.5 h in CG height 169 Kr lb/in Total rear suspension spring rate 285 Kf lb/in Total front suspension spring rate 10 e in Height of rear imaginary pivot d 33.224992 in Imaginary pivot to rear wheel e%d .30097825 Ratio e/d (if e and/or d unknown, default = h/L + h/L*Kr/Kf) thetap 9.574E-16 deg Pitch angle of vehicle

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 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 1 g's Acceleration of gravity: default = 1 g 1 ax g's Acceleration in x-direction 4074 W lb Total weight 108.5 L in Wheelbase 20.5 h in CG height 169 Kr lb/in Total rear suspension spring rate 285 Kf lb/in Total front suspension spring rate 10 e in Height of rear imaginary pivot d 52.926829 in Imaginary pivot to rear wheel .18894009 e%d Ratio e/d (if e and/or d unknown, default = h/L + h/L*Kr/Kf) thetap 1.4262419 deg Pitch angle of vehicle

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 Plot parameters: mindelta -45 deg Minimum steer angle (default = -45 deg) maxdelta 45 deg Maximum steer angle (default = +45 deg) 800 Fzl lb Vertical load on left wheel 600 Fzr lb Vertical load on right wheel 1 d in Lateral offset at ground 10 lambda deg Lateral inclination angle -45 delta deg Steer angle 5 nu deg Caster angle Moments due to vertical force: MvLatL 98.230243 lb-in from lateral inclination, left wheel MvLatR 73.672682 lb-in from lateral inclination, right wheel MvLat 171.90293 lb-in from lateral inclination, total MvCasL 49.302733 lb-in from caster, left wheel MvCasR -36.97705 lb-in from caster, right wheel MvCas 12.325683 lb-in from caster, total Mv 184.22861 lb-in 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 on the Toolbar or by pressing function key F9. The Variable Sheet should then look like the following.

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

40 V mph Velocity 500 R ft Radius of turn ac .21394745 g's Centrifugal (horizontal) acceleration 60 t in Track width (tread) 20 h in Height of CG 2200 W lb Weight of vehicle = M*g g 1 g's Acceleration of gravity (default: 1 g) M 2200 lbm Mass of vehicle .1 ay g's Acceleration along vehicle y-axis az 1.0177296 g's Acceleration along vehicle z-axis Fzi 1046.1692 lb Normal force on inside tire Fzo 1192.8359 lb Normal force on outside tire phi 6.4644744 deg Road cross-slope angle Rollover thresholds: Vroll 120.5635 mph Velocity acroll 1.9436535 g's Centrifugal acceleration ayroll 1.8187084 g's Acceleration along vehicle y-axis azroll 1.2124722 g's Acceleration along vehicle z-axis

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 V 29.109577 mph Velocity 500 R ft Radius of turn ac .11330756 g's Centrifugal (horizontal) acceleration 60 t in Track width (tread) 20 h in Height of CG 2200 W lb Weight of vehicle = M*g g 1 g's Acceleration of gravity (default: 1 g) M 2200 lbm Mass of vehicle 0 ay g's Acceleration along vehicle y-axis az 1.0063988 g's Acceleration along vehicle z-axis Fzi 1107.0387 lb Normal force on inside tire Fzo 1107.0387 lb Normal force on outside tire 6.4644744 phi deg Road cross-slope angle Rollover thresholds: Vroll 120.5635 mph Velocity acroll 1.9436535 g's Centrifugal acceleration ayroll 1.8187084 g's Acceleration along vehicle y-axis azroll 1.2124722 g's 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.