Transcript Topic12.Presentation.ICAM

NAZARIN B. NORDIN
[email protected]
What you will learn:
• Tractive effort, tractive resistance, braking
efficiency
• Tractive resistance components: rolling/
gradient/ air resistance
• Energy dissipated/ power required at constant
velocity on level plane, accelerating/ braking
forces applied on level plane, braking
efficiency
Vehicle Dynamics
CEE 320
Steve Muench
Outline
1. Resistance
a. Aerodynamic
b. Rolling
c. Grade
2.
3.
4.
5.
Tractive Effort
Acceleration
Braking Force
Stopping Sight Distance (SSD)
Main Concepts
•
•
•
•
•
Resistance
Tractive effort
Vehicle acceleration
Braking
Stopping distance
F  ma Ra  Rrl  Rg
Resistance
Resistance is defined as the force impeding
vehicle motion
What is this force?
1. Aerodynamic resistance
2. Rolling resistance
3. Grade resistance
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 
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:
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
For small angles,
sin  g  tan g
Rg  W tan g
tan g  G
Rg  WG
θg
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
Availabletractiveeffort minFe , Fmax 
Tractive Effort Relationships
Engine-Generated Tractive Effort
• Force
M e 0 d
Fe 
r
• Power
Fe = Engine generated tractive effort
reaching wheels (lb)
Me = Engine torque (ft-lb)
ε0 = Gear reduction ratio
ηd = Driveline efficiency
r = Wheel radius (ft)
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
ε0 = gear reduction ratio
Typical Torque-Power Curves
Maximum Tractive Effort
• Front Wheel Drive Vehicle
• Rear Wheel Drive Vehicle
• What about 4WD?
Fmax 
Fmax 

lr  f rl h 
W
L
h
1
L

l
W
f
 f rl h 
L
h
1
L
Diagram
θg
Vehicle Acceleration
• Governing Equation
F   R   mma
• Mass Factor
(accounts for inertia of vehicle’s rotating parts)
 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
Driveline efficiency 90%
Center of Gravity 20 inches high
Braking Force
• Front axle
Fbf
• Rear axle
max

Fbr max 
W lr  h  f rl 
L
W l f  h  f rl 
L
Braking Force
• Ratio
• Efficiency
l r  h  f rl  front
BFR 

l f  h  f rl  rear
b 
g max

Braking Distance
V
• Theoretical
– ignoring air resistance
• Practical
b
V 
S
2 g b   f rl  sin  g 
2
1
V12  V22
d
a

2 g   G 
g

d p  V1t p
• Perception
ds  d  d p
2
2
For grade = 0
V12  V22
d
2a
Stopping Sight Distance (SSD)
• Worst-case conditions
– Poor driver skills
– Low braking efficiency
– Wet pavement
• Perception-reaction time = 2.5 seconds
• Equation
2
V1
SSD 
 V1tr
a

2 g   G 
g

Stopping Sight Distance (SSD)
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
1.472
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
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
Primary References
• Mannering, F.L.; Kilareski, W.P. and Washburn, S.S. (2005). Principles of
Highway Engineering and Traffic Analysis, Third Edition). Chapter 2
• American Association of State Highway and Transportation Officals
(AASHTO). (2001). A Policy on Geometric Design of Highways and
Streets, Fourth Edition. Washington, D.C.