Tractive Effort (TE) and Tractive Resistance (TR)

Tractive effort (TE):

t
[N m]
and
TE =T
w
/ R
w
[N]
where:
T
w
is the wheel torque [N m]
T
e
is the engine torque [N m]
i
g
is the gearbox ratio
i
f
is the final drive ratio

t
is the transmission efficiency
TE is the tractive effort [N]
R
w
is the rolling radius or effective radius [m]

Maximum value of TE:
If the rear wheels have such a torque applied to them that they are on the point
of slipping in the plane of the wheel, then the value of adhesive force between
tire and road surface has been reached ( W
R
), where is the coefficient of
adhesion between tire and road surface, and W
R
is the weight on the rear
wheels. Under these conditions the engine torque has reached the maximum
useful value for the present set of conditions, as any increase would only result
in slip and lower acceleration. Static friction is higher than kinetic friction.

Maximum road speed:
When tractive effort TE is equal to the tractive resistance TR a vehicle would be
either stationary or moving at a uniform velocity. It the tractive effort exceeds
the resistance the vehicle will accelerate until the tractive effort once again
equals the tractive resistance. It follows that if the tractive resistance becomes
greater than the tractive effort, the vehicle will decelerate and come to rest
unless the tractive effort is increased or the resistance reduced.

Maximum possible road speed, [Maximum (maximum road speed)]
The maximum road speed of a vehicle in top gear on a level road is dependent
upon the rolling and air resistance, the maximum power at the road wheels and
the choice of a suitable final drive ratio.
Performance curves of resistance power (P
R
) and power at the road wheels (P
r
)
are plotted on a km/h base. The relationship of one to the other will ascertain
whether the vehicle is over geared or under geared. The optimum final drive
setting will ensure that the maximum power available P
r
crosses the resistance
power curve P
R
to give the maximum possible road speed as shown in the
figure. If the power curve P
r
is moved to the left relative to the resistance curve
P
R
the maximum road speed will be reduced, but more power becomes available
for acceleration, etc., and the vehicle is termed under geared. The vertical
distance between the two curves shows the power available for acceleration. If
the P
r
curve is moved to the right of the optimum setting, maximum road speed
will be reduced together with power available for acceleration and the vehicle is
over geared. These graphs, together with other information, are required
before the final drive ratio and gearbox ratios for the vehicle can be decided.

Data and methods for setting final drive and gearbox ratios

Final drive ratio
Figures for power at the road wheels are calculated for the road speed range,
and resistance power calculated for level road conditions using both rolling and
air resistance data. The graphs are plotted on a km/h base. The exact location of
the curves relative to each other will provide under gearing or over gearing
conditions. The point where the power available at the road wheels crosses the
resistance power denotes maximum road speed, and it will be noticed that a
slight under gearing reduces the maximum road speed by a very small amount,
as shown in the figure, but the increase in surplus tractive effort, and therefore
power, more than compensates for the speed loss by providing good
acceleration and flexibility.



Setting the final drive ratio (i
f
):
The graphs and/or data are used to determine the engine speed at maximum
road speed in the following formula for calculating the final drive.

Let us assume that the vehicle is designed to have a maximum speed of(v km/h
at direct drive.

V
max
=2 R
w
(N
w
)
max
x (60/1000) = 0.377 R
w
(N
w
)
max



and
(N
w
)
max
= v
max
/ (0.377 R
w
)

This (N
w
)
max
will occurs at the consonance engine speed at maximum engine
power (N
e
)
p max
.

The relationship between the two speeds is:

(N
w
)
max
= (N
e
)
p max
/ (i
g
i
f
),

We have i
g
= 1, since the vehicle is moving at direct drive, then;

(N
w
)
max
= (Ne)
p max
/ i
f
,

from this relation we conclude:

i
f
= (N
e
)
p max
/ (N
w
)
p max
= [(N
e
)
p max
(0.377 R )] / v
max



Worked example:
A vehicle is to have a maximum road speed of 150 km/h. If the engine develops
its peak power at 6000 rev/mm and the effective road diameter is 0.54 m,
determine the final drive gear ratio.
Solution:
i
f
= [(N
e
)
p max
(0.377 R )] / v
max


= (0.377 x (0.54 / 2) x 6000)/ 150 = 4.07:1


Settin2 the bottom sear ratio (i
g1
):
This ratio is dependent on the maximum tractive effort available in top gear, and
the maximum resistance the vehicle is likely to encounter: maximum gradient +
rolling resistance. The air resistance is neglected, since the vehicle speed to
climb the maximum gradient is too low. The greatest gradient that is likely to be
encountered is decided by the terrain the vehicle is to operate over. This is
normally means a maximum gradient of 5 to 1 and in extreme 4 to 1.

* A safety factor has to be employed which is necessary due to the fall in engine
performance during the engines working hfe, and as a safeguard in cases of
poor performance due to maladjustment of ignition or carburetion and the like.
This safety factor is usually applied after intermediate ratios have been selected
and the bottom ratio is moved out of geometrical progression.

The torque required at the wheel (T
w
) is equal to the resistance torque (T
TR
), AR
is neglected at low car velocity.

T
w
= T
TR
= (RR + GR ) x R
w


The torque at the wheel in this case is equal to:

Then;
T
w
= T
max
i
g
i
f

t

T
max
i
g
i
f

t
= (RR+GR) R
w

i
g
= [(RR + GR) R
w
] / (T
max
i
f

t
)
where:
RR is the rolling resistance
GR is the maximum gradient
resistance the car can climb
R
w
is the effective tire radius

t
is the total transmission
efficiency at the bottom gear
i
f
is the final drive ratio
(obtained from the previous
section)


Worked example:
A vehicle weighting 1500 kg has a coefficient of rolling resistance of 0.015.
The transmission has a final drive ratio 4.07: 1 and an overall mechanical
efficiency of 85%.
If the engine develops a maximum torque of 100 Nm and the effective road
wheel radius is 0.27 m, determine the gearbox bottom gear ratio, if the vehicle
to climb a maximum gradient of 4:1.
Solution:
RR = f
r
W =f
r
mg = 0.015 x 1500 x 9.81 = 220.725 N

GR = W sin =(mg) sin = 1500 x 9.81 x (1/4) = 3678.75 N

TR = RR + GR = 220.725 + 3678.75 = 3899.475 N

i
g
= [(RR + GR) R
w
] / ( T
max
i
f

t
)

i
g
= [3899.475 x 0.27] / (100 x 4.07 x 0.85) = 3.04: 1


Setting intermediate sear ratios
Ratio between top and bottom gears should be spaced in such a way that they
will provide the tractive effort-speed characteristics as close to the ideal as
possible. Intermediate ratios can be best selected as a first approximation by
using a geometric progression. This method of obtaining the gear ratios requires
the engine to operate within the same speed range in each gear, which is
normally selected to provide the best fuel economy.

Before the intermediate ratios can be calculated a constant called the engine
speed ratio has to be found. This is the ratio of engine re/mm at which
maximum torque is generated to the rev/mm when maximum engine power is
produced.

Engine speed ratio (K) = N
e
at maximum torque / N
e
at maximum power

The working engine speed range can be shown graphically and the larger the
distance between point of maximum torque and point of maximum power
rev/min, the greater the flexibility as seen in the figure. The engine with greater
flexibility (engine B) needs a gearbox with fewer gear ratios.

* The engine working range (between maximum torque and maximum power)
is suitable for car driving condition, when the car encounter additional minor
resistance while moving, that will lead to decrease in car velocity (decrease in
engine rpm) the reduction in engine rpm will lead to increase in the engine
torque that will overcome this resistance, without the need to change to a lower
gear.


The engine speed ratio K is used to assist in the setting of the intermediate ratios
once bottom gear has been calculated. This constant will provide ratios in a
geometrical progression. This can be written as:

i
g2
/i
g1
= i
g3
/i
g2
= i
g4
/i
g3
= i
g5
/i
g4
= N
L
/ N
H
= K

where:
i is the n gearbox ratio
N
L
is the lower engine speed rev/mm,
N
H
is the highest engine speed

i
g5
/i
gl
= (i
g2
/i
g1
) (i
g3
/i
g2
) (i
g4
/i
g3
)

(i
g5
/i
g4
) = (K) (K) (K) (K) = K
4
= K
(n-1)


Where n is the number of gear shill. The gear ratios can be obtained from the
above equation as follow:

i
g1
= i
g1
K
(1-1)
= i
g1
K
0
= i
g1
(the bottom gear ratio)
i
g2
= i
g1
K
(2-1)
= i
g1
K
i
g3
= i
g1
K
(3-1)
= i
g1
K
2

i
g1
= i
g1
K
(4-1)
= i
g1
K
3

i
g1
= i
g1
K
(5-1)
= i
g1
K
4


For commercial vehicles, the gear ratios in the gearbox are often arranged in
geometric progression. For passenger cars, to suit the changing traffic condition,
the steps between the ratios of the upper two gears is often closer than that
based on geometric progression. As a result, this will affect the section of the
lower gears to some extent. In case the power to weight ratio is decreased, the
number of gearbox shifts need to be increase, so we can get more surplus
tractive effort to the acceleration.
After the intermediate ratios have been decided upon a safety factor may be
placed on the bottom gear lowering it s ratio wise out of geometrical
progression, and any manufacturing and design modifications can be applied.

Sitting intermediate gear ratios graphically:
Consider the engine to vehicle speed characteristics for each gear ratio as shown
in the figure. When changing gear the engine speed will drop from the highest
N
H
to the lowest N
L
without any change in the road speed i.e. v
1
, v
2
, etc.

Worked example:
A transmission system for a vehicle is to have an overall bottom and top gear
ratio of 20:1 and 4.8 respectively. If the minimum and maximum speeds at each
gear changes are 2100 and 3000 rev/mm respectively, determine the following:
a) the intermediate overall gear ratios b) the intermediate gear box and top gear
ratios.

K=N
L
/N
H
= 2100 /3000 = 0.7

a) Total 1
st
gear ratio = (i
g1
i
f
) = 20: 1
Total 2
nd
gear ratio = (i
g1
i
f
) K = 20 x 0.7 = 14.0: 1
Total 3
rd
gear ratio = (i
g1
i
f
) K
2
= 20 x 0.7
2
= 9.8:1
Total 4
th
gear ratio = (i
g1
i
f
) K
3
= 20 x 0.7
3
= 6.86: 1
Total 5
th
gear ratio = (i
g1
i
f
) K
4
= 20 x 0.7
4
= 4.8: 1

i
g1
= Total 1
st
gear ratio/i
f
= 20/4.8 = 4.166:1 (the bottom gear ratio)
i
g2
= Total 2
nd
gear ratio/i
f
= 14/4.8 = 2.916:1
i
g3
= Total 1
st
gear ratio/i
f
= 9.8/4.8 = 2.042:1
i
g4
= Total 1
st
gear ratio/i
f
= 6.86/4.8 = 1.429:1
i
g5
= Total 1
st
gear ratio/i
f
= 4.8/4.8 = 1:1 (top gearbox ratio and direct
drive)



Setting the Gearbox and Final Drive Ratios

Sitting intermediate gear ratios graphically:
Consider the engine to vehicle speed characteristics for each gear ratio as shown
in the figure. When changing gear the engine speed will drop from the highest
N
H
to the lowest N
L
without any change in the road speed i.e. v
1
, v
2
, etc.


In the figure below a method of showing the engines speed ratio is given. This
is based on the assumption that engine revolutions should not fall below the
speed of maximum torque. The point at which the top gear line crosses the
maximum torque line is the point at which a gear change should be mad. The
change is indicated by a vertical line ab, and the line drawn from b to the origin
0 represents the conditions after the change down. The process is repeated for
the other ratios. The horizontal distance between the vertical lines increase in
geometrical progression or (approximately to this) depending upon whether
modifications to some or all the calculated ratios has been made.


top gear = 3200/3200 = 1 to 1
3
rd
gear = 3200/ 1800 = 1.77 to 1
2
nd
gear = 3200/1100 =2.91 to 1
1
st
gear = 3200/740 = 4.325 to 1
Engine speed ratio = 1800 / 3200 = 0.5625

* Note that in the above gear settings, after doing the calculations the first gear
has been lowered slightly out of geometrical progression with second, and third
gear moved closer to top gear, hence the speed ratio 0.5625 would not now
apply.


Worked example (calculating the possible gear ratios for a vehicle):

Calculate the gear and final drive ratios for a vehicle from the given data:
Maximum engine torque is 105.2 N m at 2100 rev/mm.
Maximum power is 37.3 kW at 4000 rev/mm,
giving a road speed of 130 km/h obtained from graphs and data.
The rolling radius of road wheels is 0.366 m.
Transmission efficiency is 90%
and the maximum tractive resistance is 4890 N.

V = 0.377 R
w
N
w
= 0.377 R
w
N
e
/ (i
g
i
f
)

V
max
= 0.377 R
w
N
w max
= 0.377 R
w
N
e(pmax)
/( i
g
i
f
)

i
f
= 0.377 R
w
N
e(pmax)
/( i
g
V
max
) = 0.377 x 0.366 x 4000/(1 x 130) = 4.245 : 1
(final drive)

* maximum tractive resistance (TR
max
) = maximum tractive effort at bottom
gear

TR
max
= T
e max
i
g
i
f

t
/ R
w


i
g
= TR
max
R
w
/ (T
e max
i
f

t
) = 4895 x 0.366 / (105.2 x 4.245 x 0.9) = 4.458 : 1

* engine ratio (k) = N
e (max torque)
/ N
e (max power)
= 2100/4000 = 0.525

i
g1
= 4.458 :1

i
g2
= K i
g1
= 0.525 x 4.458 = 2.34 : 1

i
g3
= K
2
i
g1
= 0.525
2
x 4.458 = 1.228 : 1

i
g4
= K
3
i
g1
= 0.525
3
x 4.458 = 0.645 : 1 1 : 1



v