9/4/2009
Available Tractive Effort
Road Vehicle Performance:
Tractive Effort and Acceleration

Tractive effort determined by:

Force of vehicle’s engine

Maximum value transferable
CE 322
Transportation Engineering
Dr. Ahmed Abdel-Rahim
Maximum Tractive Effort


Maximum Tractive Effort
Define the point at which additional
engine-generated tractive effort is not
productive.
Fig. 2.3
Examine a free body diagram




L = wheelbase
h = height of the center of gravity
lf, lr = distance from the front, rear axle to the CG
Wf, Wr = weight of vehicle on front, rear axle
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9/4/2009
Maximum Tractive Effort
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Maximum Tractive Effort
Examine the normal load on the drive axle.

Assuming a rear-wheel drive car, determine the
normal load on the rear axle:
Rearranging terms, assuming
cos g = 1, and then substituting into
Eq. 2.2 ( F = ma + Ra + Rrl + Rg ), yields:
Wr 
Wr 
Ra h  Wl f cos  g  mah  Wh sin  g
lf
L
W
h
F  Rrl 
L
Eq. 2.11
Eq. 2.10
L
Grade moment: + for incline
Maximum Tractive Effort

And from basic physics we know
(for a rear-wheel-drive car):
Fmax  Wr
 = coefficient of road adhesion
Maximum Tractive Effort

Substituting Eq. 2.11 into Eq. 2.12 yields:
Fmax 
Eq. 2.12

W l f  f rl h  / L
1  h / L
Eq. 2.14
Similarly, we have the following formula for a
front-wheel-drive vehicle:
Fmax 
W lr  f rl h  / L
1  h / L
Eq. 2.15
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Engine-Generated
Tractive Effort (Fe)

Measures of engine output:





torque (Me)
power (hpe)
Torque = work generated by engine

Engine-Generated
Tractive Effort (Fe)
the twisting moment and
expressed in foot-pounds (ft-lb).
Power = engine work per unit of time:
hp e =
2πM e ne
550
Pe =
2 M e ne
1000
Eq. 2.16
hpe = engine-generated horsepower (1 horsepower equals
550 ft-lb/s),
Pe = engine-generated power in kW,
Me = engine torque in ft-lb (N-m), and
ne = engine speed in crankshaft revolutions per second.
Engine-Generated
Tractive Effort (Fe)
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Typical torque-power curve for a gasoline-powered
engine
Engine-Generated
Tractive Effort (Fe)
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Tractive effort needed


Torque available

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greatest at lower vehicle speeds
Greatest at higher engine speeds.
What is the problem here?
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Engine-Generated
Tractive Effort (Fe)
Engine-Generated
Tractive Effort (Fe)
V
2re (1  i)
Fig. 2.5

0
Some tractive effort is lost

Sources of loss: transmission, differential

Tractive effort lost in driveline: 5-25%

mechanical efficiency term, d
V = vehicle speed (fps),
r = drive wheel radius (ft),
ne = engine speed in
crankshaft
revolutions per
second,
εo = overall gear ratio, and
i = slippage of drive axle.
Engine-Generated
Tractive Effort (Fe)

Tractive effort (0) is reduced

engine crankshaft rev’s / drive wheel rev’s.
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If 0 is high then what is max speed?
V

What does an 0 = 3 mean?
2re (1  i)
0
How does this relate to Fe?
Engine-Generated
Tractive Effort (Fe)
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Engine-generated tractive effort
reaching the drive wheel:
Fe 
M e 0d
r
Eq. 2.17
Fe = engine-generated tractive effort in lb (N)
Me = engine torque in ft-lb (N-m)
r = radius of the drive wheels in ft (m)
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Engine-Generated
Tractive Effort (Fe)

Tractive Effort (Fe)
The relationship between vehicle speed
and engine speed is:
V
2rne (1  i)
0

Eq. 2.18
V = vehicle speed in ft/s (m/s)
ne = engine speed in rev/s
i = driveline slippage (generally taken as
2-5% for passenger cars)
Vehicle Acceleration
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Eq. 2.2 can be used again, with an
additional term added
m = mass factor
 inertia of rotating parts
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Minimimum [Fmax, Fe]
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F = ma + Ra + Rrl + Rg (Eq. 2.2)

What happens if…
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F = Ra + Rrl + Rg
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F < Ra + Rrl + Rg

F > Ra + Rrl + Rg
Vehicle Acceleration

Rearranging Eq. 2.2 with mass factor
F   R  γ m ma
F = mma + Ra + Rrl + Rg

Available tractive effort (F)

Eq. 2.19
Approximation of mass factor
γm = 1.04  0.0025ε02
Eq. 2.20
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Vehicle Acceleration

Force available to accelerate:
Fnet  F   R
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Vehicle Acceleration
Fig. 2.6
Fnet = 0, vehicle at its top speed
Relationship between Fnet, F (lesser of Fmax
and Fe) and R is shown in Fig. 2.6
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