This document contains lecture slides for a CEE320 Vehicle Dynamics course. It covers topics related to vehicle resistance forces like aerodynamic and rolling resistance. It also discusses vehicle tractive effort, acceleration, braking forces, and stopping sight distance. Equations are provided for concepts like available tractive effort, maximum acceleration, braking distance, and quick estimation of stopping sight distance. Examples are included, such as calculating maximum acceleration for a 1989 Ford Mustang. Primary references for further information are cited.
5. CEE320
Spring2008
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%)
2
2
VACR fDa
ρ
=
3
2
VACP fDRa
ρ
=
sec
5501
lbft
hp
⋅
=
from National Research Council Canada
6. CEE320
Spring2008
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:
WfR rlrl =
+=
147
101.0
V
frlWVfP rlrlR =
sec
5501
lbft
hp
⋅
=
8. CEE320
Spring2008
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
( )max,minefforttractiveAvailable FFe=
13. CEE320
Spring2008
Maximum Tractive Effort
• Front Wheel Drive Vehicle
• Rear Wheel Drive Vehicle
• What about 4WD?
( )
L
h
L
hfl
W
F
rlf
µ
µ
−
−
=
1
max
( )
L
h
L
hfl
W
F
rlr
µ
µ
+
+
=
1
max
16. CEE320
Spring2008
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
23. CEE320
Spring2008
SSD – Quick and Dirty
( )
( ) ( )
( ) a
VV
V
V
Ggag
VV
d
22
2
22
1
2
2
2
1
075.1
2.11
075.1
2.11
1
2
47.1
02.322.112.322
047.1
2
==××=
+×
−×
=
±
−
=
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)
ppp VttVd 47.147.1 1 =××=
V = V1 in mph
a = deceleration, 11.2 ft/s2
in US customary units
tp = Conservative perception / reaction time = 2.5 seconds
ps Vt
a
V
d 47.1075.1
2
+=
25. CEE320
Spring2008
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.
Notas do Editor
Power is in ft-lb/sec
Rolling resistance = 2 components
Hysteresis = energy loss due to deformation of the tire
Adhesion = bonding between tire and roadway
Low profile tires reduce r and increase tractive effort
Torque and HP always cross at 5252 RPM. Why? Look at the equation for HP
Torque determines acceleration – a car will accelerate at its maximum when torque to the drive wheels is maximum (this is lower than engine output torque due to losses within the system). Max acceleration at max torque.
However, as you accelerate you need to up-shift and change the gearing ratio. When you shift, you change the RPM to the drive axle(s) to a lower value. This, in turn, lowers torque. Therefore, as you accelerate, the longer you can stay in a higher gear the less loss in torque overall. One way to express this is through horsepower because horsepower takes into account engine speed.
A higher horsepower car can stay in a lower gear longer and thus suffer less torque loss.
It is better to make torque at high RPM than low RPM because you can take advantage of gearing.
In theory, different gear ratios - most commonly four or five in cars' gearboxes - should mask different torque characteristics by altering engine speed to suit but the reality is that engines which produce high torque figures at low revolutions respond much more readily in give and take driving.The practical advantages come in the form of reduced gear changing, lower engine revs and wear and, invariably, lower fuel consumption in all conditions other than constant speed driving.
For 4WD
Fmax = μW (if your 4WD distributes power to ensure wheels don’t slip, which is common)
For a front wheel drive car, sum moments about the rear tire contact point:
-Rah – Wsinθh + Wcosθlr + mah - WfL = 0
cosθ = about 1 for small angles encountered
-Rah – Wsinθh + Wlr + mah - WfL = 0
WfL = -Rah – Wsinθh + Wlr + mah
WfL = + Wlr – Wsinθh – Rah + mah
Wf = (lr/L)W + (h/L)(-Wsinθ – Ra + ma)
But… Wsinθ = Rg
Substituting: Wf = (lr/L)W + (h/L)(-Rg – Ra + ma)
We know that… F = ma + Ra + Rrl + Rg Therefore, -F + Rrl = -ma – Ra– Rg
Wf = (lr/L)W + (h/L)(-F + Rrl)
Now, Fmax = μWf and Rrl = frlW
Substituting: Fmax = μ((lr/L)W + (h/L)(-Fmax + frlW))
Simplifying: Fmax + (μh/L)Fmax = μ((lr/L)W + (h/L)(frlW))
Fmax(1 + μh/L) =( μW/L)((lr + hfrl)
Tire size
P = passenger car
1st number = tire section width (sidewall to sidewall) in mm
2nd number = aspect ratio (sidewall height to width) in tenths (e.g. 60 = 0.60)
3rd number = wheel diameter
Practical comes from V22 = V12 + 2ad (basic physics equation or rectilinear motion)
a = 11.2 ft/sec2 is the assumption
This is conservative and used by AASHTO
Is equal to 0.35 g’s of deceleration (11.2/32.2)
Is equal to braking efficiency x coefficient of road adhesion
γb = 1.04 usually