Transcript Vehicle Dynamics CEE 320 Steve Muench 06

CEE 320
Winter 2006
Vehicle Dynamics
CEE 320
Steve Muench
Outline
1. Resistance
a. Aerodynamic
b. Rolling
c. Grade
CEE 320
Winter 2006
2.
3.
4.
5.
Tractive Effort
Acceleration
Braking Force
Stopping Sight Distance (SSD)
Main Concepts
•
•
•
•
•
Resistance
Tractive effort
Vehicle acceleration
Braking
Stopping distance
CEE 320
Winter 2006
F  ma  Ra  Rrl  Rg
Resistance
Resistance is defined as the force impeding
vehicle motion
1.
2.
3.
4.
What is this force?
Aerodynamic resistance
Rolling resistance
Grade resistance
CEE 320
Winter 2006
F  ma  Ra  Rrl  Rg
Aerodynamic Resistance Ra
Composed of:
1. Turbulent air flow around vehicle body (85%)
2. Friction of air over vehicle body (12%)
3. Vehicle component resistance, from radiators
and air vents (3%)
Ra 
CEE 320
Winter 2006
PRa 
from National Research Council Canada

2

2
CD Af V 2
CD Af V
1 hp  550
ft  lb
sec
3
Rolling Resistance Rrl
Composed primarily of
1. Resistance from tire deformation (90%)
2. Tire penetration and surface compression ( 4%)
3. Tire slippage and air circulation around wheel ( 6%)
4. Wide range of factors affect total rolling resistance
5. Simplifying approximation:
CEE 320
Winter 2006
Rrl  f rlW
PR rl  f rlWV
1 hp  550
ft  lb
sec
V 

f rl  0.011 

 147 
Grade Resistance Rg
Composed of
– Gravitational force acting on the vehicle
Rg  W sin  g
θg
For small angles, sin  g  tan  g
Rg  W tan  g
CEE 320
Winter 2006
tan  g  G
Rg  WG
Rg
θg
W
Available Tractive Effort
The minimum of:
1. Force generated by the engine, Fe
2. Maximum value that is a function of the
vehicle’s weight distribution and road-tire
interaction, Fmax
CEE 320
Winter 2006
Available tractive effort  min Fe , Fmax 
CEE 320
Winter 2006
Tractive Effort Relationships
Engine-Generated Tractive Effort
• Force
M e 0 d
Fe 
r
Fe = Engine generated tractive effort
reaching wheels (lb)
Me = Engine torque (ft-lb)
ε0 = Gear reduction ratio
ηd = Driveline efficiency
r = Wheel radius (ft)
CEE 320
Winter 2006
• Power
ft  lb  torque ft  lb  engine rpm

hp  550

 2

sec 
550
 sec 

60

 min 
Vehicle Speed vs. Engine Speed
V
2rne 1  i 
0
V = velocity (ft/s)
r = wheel radius (ft)
ne = crankshaft rps
i = driveline slippage
CEE 320
Winter 2006
ε0 = gear reduction ratio
CEE 320
Winter 2006
Typical Torque-Power Curves
Maximum Tractive Effort
• Front Wheel Drive Vehicle Fmax 
CEE 320
Winter 2006
• Rear Wheel Drive Vehicle Fmax 
• What about 4WD?

lr  f rl h 
W
L
h
1
L

l
W
f
 f rl h 
L
h
1
L
CEE 320
Winter 2006
Diagram
θg
Vehicle Acceleration
• Governing Equation
F   R   m ma
• Mass Factor
(accounts for inertia of vehicle’s rotating parts)
CEE 320
Winter 2006
 m  1.04  0.0025 02
Example
A 1989 Ford 5.0L Mustang Convertible starts on a flat grade from a dead
stop as fast as possible. What’s the maximum acceleration it can achieve
before spinning its wheels? μ = 0.40 (wet, bad pavement)
1989 Ford 5.0L Mustang Convertible
Torque 300 @ 3200 rpm
Curb Weight 3640
Weight Distribution Front 57%
Rear 43%
Wheelbase 100.5 in
Tire Size P225/60R15
Gear Reduction Ratio 3.8
CEE 320
Winter 2006
Driveline efficiency 90%
Center of Gravity 20 inches high
Braking Force
• Front axle
CEE 320
Winter 2006
• Rear axle
Fbf
max

Fbr max 
W lr  h  f rl 
L
W l f  h  f rl 
L
Braking Force
• Ratio
l r  h  f rl  front
BFR 

l f  h  f rl  rear
CEE 320
Winter 2006
• Efficiency  b 
g max

Braking Distance
• Theoretical
– ignoring air resistance
• Practical
 b V12  V22 
S
2 g b   f rl  sin  g 
V12  V22
d
a

2 g   G 
g

CEE 320
Winter 2006
• Perception d p  V1t p
• Total
ds  d  d p
For grade = 0
V12  V22
d
2a
Stopping Sight Distance (SSD)
• Worst-case conditions
– Poor driver skills
– Low braking efficiency
– Wet pavement
CEE 320
Winter 2006
• Perception-reaction time = 2.5 seconds
• Equation
V12
SSD 
 V1t r
a

2 g   G 
g

Stopping Sight Distance (SSD)
CEE 320
Winter 2006
from ASSHTO A Policy on Geometric Design of Highways and Streets, 2001
Note: this table assumes level grade (G = 0)
SSD – Quick and Dirty
1. Acceleration due to gravity, g = 32.2 ft/sec2
2. There are 1.47 ft/sec per mph
3. Assume G = 0 (flat grade)


V12  V22
1.47  V12  0
1.47 2
1
V2
V2
2
d



 V  1.075
 1.075
2 g a g  G  2  32.211.2 32.2  0
2
11.2
11.2
a
d p  1.47  V1  t p  1.47Vt p
CEE 320
Winter 2006
V2
d s  1.075
 1.47Vt p
a
V = V1 in mph
a = deceleration, 11.2 ft/s2 in US customary units
tp = Conservative perception / reaction time = 2.5 seconds
CEE 320
Winter 2006
Primary References
• Mannering, F.L.; Kilareski, W.P. and Washburn, S.S. (2005).
Principles of Highway Engineering and Traffic Analysis, Third
Edition). Chapter 2
CEE 320
Winter 2006
• American Association of State Highway and Transportation
Officals (AASHTO). (2001). A Policy on Geometric Design of
Highways and Streets, Fourth Edition. Washington, D.C.