ERRATA AND CLARIFICATIONS Fundamentals of Vehicle ...

ERRATA AND CLARIFICATIONS Fundamentals of Vehicle Dynamics by Thomas D. Gillespie Page v, Acknowledgements: Typographical error on next to the last line. The fourth name is supposed to read Chuck Houser. Page 17: Correction to Equation for FWD at bottom of the page (Subscript numeral one has been added to the “h” in the second term of the denominator) c -ζd f L L Lt θ=µ h h h -h 1+ µ 1 + ζ (1+ µ 2 + µ d 2 3 ) L L L Lt Page 20: Correct Reference 17 (incorrect author’s initials) to read as follows: Wong. J. Y., Theory of Ground Vehicles, John Wiley & Sons, New York, 1978, 330 p. Page 24: Correct Figure 2.2 (Mislabeled vertical axis). .30

10% Passenger Car, 40 lb/hp

.25 a x .20 g .15 .10

Typical Heavy Truck, 250 lb/hp

.05 0

10

20

30 40 Speed (mph)

50

60

Page 27: Add Reference 9 to the paragraph on discussion of mass factor at center of the page. The combination of the two masses is an "effective mass," and the ratio of (M + Mr)/M is the "mass factor." The mass factor will depend on the operating gear, with typical values as below [9]. Ref: 9 Taborek, J. J., Mechanics of Vehicles, Towmotor Corporation, Cleveland, Ohio, 1957, 93 p.

Page 27: Add Reference 9 to the paragraph following the table, and correct Equation (212) by adding Ntf in the second term. A representative number is often taken as [4,9]: Mass Factor = 1 + 0.04 N tf + 0.0025 Ntf2

(2-12)

Page 29: Modifications to Figure 2.5 to clarify its interpretation. 100

2.0

80

1.5

60

1.0

40

0.5

20

Efficiency (%)

Output/Input Torque Ratio

Lockup

0 0.2 0.4 0.6 0.8 1.0 Output/Input Speed Ratio

0 0

Page 32: Correct Figure 2.9 (Mislabeled horizontal axis)

Brake Mean Effective Pressure (psi)

120 Consumption (lb/bhp h) .58 .54 .58

.54

Wide Open Throttle

100 .50 .46

80

.54

60 Operating Schedule of Ideal Transmission

40

.62 .68 .72 .84 .92

20 0 500

.58

1000

1500 2000 Engine Speed (RPM)

2500

3000

Page 63: Modification to Figure 3.8. The lower expression for “Slope” has been changed to clarify its interpretation. µ pW fs 1 - µ p h/L

Front and Rear Lockup ckup

Front Lo

2000

Slope = 20

µ ph/L 1 - µp h/L

fp s sp

Slope =

er

Front Brake Force (lb)

el ec D at

1500

1 + µp h/L

io n

- µp h/L

500

Pro

por

tion

ing

Line

p

u Rear Lock

1000

µp Wrs 1 + µp h/L 500

1000

1500

2000

Rear Brake Force (lb)

Page 150: Correction to Equation (5-16). The first symbol in the numerator is the Greek letter Chi (χ), not Mu (µ). [χ ω 4 -(K 1 + K 2) ω 2]+ j [C ω 3] Z = Fb/M [χω 4 -(K 1 + K 2 χ + K 2) ω 2 + K 1 K2]+ j [K1 C ω -(1 + χ)C ω 3] Page 178: Example Problem—Ride rates and loads are per wheel. I.e., Front ride rate = 127 lb/in/wheel

Front tire load = 957 lb/wheel

Rear ride rate = 92.3 lb/in/wheel

Rear tire load = 730 lb/wheel

Page 205: Correct units on Equation (6-20) are “g/deg” not “deg/sec” as shown. Page 224: In Equation (6-65) the last of the three terms should be preceded by a minus (-) sign, rather than a plus (+) sign.

Page 230: Correction to Equation (6-73) K=

∂δ - 57.3 L ∂a y V2

(6-73)

Page 233: Correction to the equation in the middle of the page. c' = L

C αf 232 = 8.38 ft = 4.55 ft C αf + C αr 232 + 195

Page 233: Example Problem 2, part b, correct text to read: b) If the sprung mass is 2750 lb at a CG height of 10 inches above.... Page 235: Correct Reference 10 to read: Ellis, J. R., Vehicle Dynamics, Business Books Limited, London, 1969, 243 p. Page 288: Correction to Figure 8.12 to fix mislabeled wheel loads.

Steering Torque (in-lb)

200

STEERING TORQUE FROM LATERAL INCLINATION ANGLE 1" Offset 10° Inclination Angle

100

0 Left Wheel (600 lb) Right Wheel (800 lb)

-100

Total -200 -45

-30

-15 0 15 Steering Angle (deg)

30

45

Page 289: Correct Equation (8-5) by adding minus (-) sign to the right of equals sign. ML = - (F yl + F yr) r tan ν

(8−5)

Lateral Acceleration, a y

Page 313: Correction to Figure 9.2

Rollover Threshold

Un

sta

ble t φ = tan-12 h

Roll Angle, φ Page 313: Correction on last line of the page. ...equilibrium lateral acceleration reaches zero (φ = tan-1 (t/2h)).

Lateral Acceleration, ay

Page 317: Correction to Figure 9.4

Rφ

Rollover Threshold (lift-off of second axle)

Lift-off of first axle

Roll Angle, φ

t φ = tan-1 2 h