[David Crolla_Behrooz Mashadi]Vehicle Powertrain Systems-Integration and Optimization_Solution Manual

Problem 2.1

Use the MATLAB program of Example 2.3.3 to study the effects of changing engine parameters

on its torque generation performance.

a) Find the effect of a 10% reduction of piston and connecting rod masses on the engine torque.

b) Find the effect of reducing the connecting rod length by 10%.

c) Find the effect of reducing the connecting rod inertia by 10%.

Solution:

The solution is straightforward. Inside the program listing, just change the numerical

values and run the program.

a) Reducing the piston mass by 10% gives mP=387 g. The average engine torque will

not change and remain at 37.81 Nm. The same result is obtained for the connecting

rod mass with mC=396 g. Also it can be observed that for even larger changes, the

average engine torque does not change.

The maximum engine torque, however, increases 1.5% from 493.2 to 500.2 Nm

when reducing the piston mass by 10%. It increases by 0.6% when reducing the

connecting rod mass by 10%.

b) Reducing the connecting rod length by 10% gives l=126 mm. The average engine

torque will increase to 38.35 Nm, namely by 1.5%. The increase of maximum torque

for this case is more than 2%.

c) Reducing the connecting rod inertia by 10% gives IC=0.00135 kgm2. The average

engine torque will remain unchanged for this change. The maximum torque in this

case is 492.2 Nm, i.e. 0.2% decrease.

It is, therefore, concluded that the connecting rod length is the only influential factor

on the average engine torque.

It should also be noted that the average engine speed was assumed to remain at the

same value of 3000 rpm. In other words, the effect of engine speed itself was not

studied.

Problem 2.2

Use the data of Problem 2.1 to study the effects of changing the engine parameters on the engine

bearing loads.

Solution:

The engine bearing loads include the main bearing and the crank pin bearing loads.

The latter was already included in the program. To include the former, the following

statements should be included at the end of the program listings:

% Main bearing force

mB=lA*mC/l;

FICW=mB*R*omeg^2; % The inertia force of the counterweight is considered to balance that of mB

for i=1: 361

thetai=theta(i)*pi/180;

FBx=Bx(i)+FICW*cos(thetai);

FBy=By(i)+FICW*sin(thetai);

FB(i)=sqrt(FBx^2+FBy^2);

end

figure

plot(theta, FB/1000)

xlabel('Crank angle (deg)')

ylabel(Main bearing force (kN))

The overall crank-pin bearing force also is:

B=sqrt(Bx.^2+By.^2);

Now the variation of bearing forces can be studied. To this end it is useful to disable

all plot statements other than for the bearing forces.

a) Reducing the piston mass by 10% changes the average and maximum of the main

bearing force FB from 4.510 and 14.971 kN to 4.410 and 15.207 kN (1.5% reduction

and 1.6% increase respectively). The average and maximum of the crank-pin bearing

force B changes from 4.224 and 14.524 kN to 4.155 and 14.760 kN (2% reduction

and 1.6% increase respectively). A 10% reduction in the connecting rod mass

reduces the averages of main bearing and crank-pin forces by 3.3% and 2.5%, while

increasing the maximum of the same forces by 1% and 1.3% respectively.

b) Reducing the connecting rod length by 10% changes the averages of the main

bearing and the crank-pin bearing forces to 4.538 and 4.276 kN (i.e. 0.6% and 1.2%),

and at the same time reduces the maximum of the forces to 14.897 and 14.497 kN

(i.e. 0.5% and 0.2% each).

c) Reducing the connecting rod inertia by 10% results in the average and maximum

of the main bearing force of 4.496 and 14.969 kN (i.e. -0.3% and -0.01%). The

average and maximum of the crank-pin bearing force become 4.210 and 14.521 kN

respectively (i.e. -0.3% and 0.02%).

Therefore, the important parameters are the connecting rod mass and then the piston

mass. The connecting rod inertia has little effect on the bearing loads.

Problem 2.3

Derive expressions for the gudgeon-pin and crank-pin bearing forces A and B of the simplified

model according to the directions of Figure 2.32.

Results: tanyx AA , PAPPy ammFA )( , xx AB and yy AB .

Solution:

According to the FBD shown in Figure S2.3 (a), one simply writes:

Wx FA

PAPPy ammFA )(

Recalling that in the simplified model the forces at A and B are in the direction of the

link AB, then:

tanyx AA

and from Figure S2.3 (b),

xx AB and yy AB

The resultant forces are cos

)(22 PAPP

yx

ammFAA

BA

The results show that only information on the geometry, the pressure force (FP) and

the piston acceleration (aP) are needed to obtain the bearing forces.

Figure S2.3 Free body diagrams for the simplified mode

Problem 2.4

For the engine of Example 2.3.3 compare the gudgeon-pin and crank-pin resultant forces of the

exact and simplified engine models at 3000 rpm.

Hint: To find the gudgeon-pin forces of the exact model use Equations 2.80, 81, 84 and 85.

Ax

Ay

FP

FW

FIP

A

B

Ay

Ax

Bx

By

(a) (b)

A

Solution:

The crank-pin forces were calculated in Example 2.3.3 for the exact model. Those of

the simplified model are Bx and By of Problem 2.3. The gudgeon-pin forces of the

exact model from Equations 2.80, 2.81 and 2.85 are:

xIxx BFA , PIPy FFA

where Bx is given in Equation 2.84. The gudgeon-pin forces of the simplified model

are Ax and Ay of Problem 2.3. The following simple MATLAB program can be used

to evaluate the bearing forces.

for i=1: 361

beta=asin(Rl*sin(theta(i)*pi/180));

Bx(i)=(-FIP(i)-Fp(i)+lB*FIy(i)/l)*tan(beta)-lA*FIx(i)/l-TIG(i)/l/cos(beta);

Ay_s(i)=-FPt(i);

Ax_s(i)=-Ay_s(i)*tan(beta);

end

By=Fp+FIP-FIy;

Ay_e=-FIP-Fp;

Ax_e=-Bx-FIx;

Bx_s=-Ax_s;

By_s=-Ay_s;

A_s=sqrt(Ax_s.^2+Ay_s.^2);

A_e=sqrt(Ax_e.^2+Ay_e.^2);

B_s=sqrt(Bx_s.^2+By_s.^2);

B_e=sqrt(Bx.^2+By.^2);

The results are shown in Figures S2.4a and S2.4b.

Figure S2.4a

Figure S2.4b

Problem 2.5

Show that for the exact engine model the average of term Te-FWh during one complete cycle

vanishes.

0 90 180 270 360 450 540 630 7200

2000

4000

6000

8000

10000

12000

14000

16000

18000

Crank angle (deg)

Cra

nk-p

in b

eari

ng

fo

rce

s (

N)

Exact

Simplified

0 90 180 270 360 450 540 630 7200

2000

4000

6000

8000

10000

12000

14000

16000

18000

Crank angle (deg)

Gu

dge

on

-pin

bea

ring

fo

rce

s (N

)

Exact

Simplified

Figure S2.5 The variation of Te-FWh term with crank angle

Problem 2.6

Construct the firing map for a three cylinder in-line engine with cranks at 0-120-240 degrees.

0 90 180 270 360 450 540 630 720-8

-6

-4

-2

0

2

4

6

8

Crank angle (deg)

Torq

ue

te

rm (

Nm

)

Solution:

The term Te-FWh for the simplified model is always zero (Equation 2.103). For the

exact model, however, from Equation 2.90 it is:

ICIGAIyAIxWe TTlFlhFhFT sin)cos(

A closed form solution that deals with the integration of the above equation over the

complete cycle, is quite complicated. Within the MATLAB program, however, the

torque term can be determined easily as shown in Figure S2.5 for one complete

cycle. It is clear that the average of the function is zero.

Figure S2.6 Firing order chart for Problem 2.6

Problem 2.7

Construct the firing map for a 4 cylinder 60 V engine with cranks at 0-0-60-60 degrees.

1

2

3

1

Relativ

e state of stro

kes

Exhaust

Cran

k lay

ou

t

Stroke order

Firing order

0

120

240

180 180 180

Exhaust Intake

2 1

2

3

3

Compression

Exhaust Intake Compression

Compression

Intake

240

180

120

60

240

Solution:

The solution is very similar to that of in-line six cylinder engine of Figure 2.55. The

difference is due to exchange of orders of the cylinders 2 and 3. The result for the 1-

2-3 firing order is depicted in Figure S2.6. A 1-3-2 firing order is also possible by

changing the stroke orders of cylinders 2 and 3.

1

3 2

4

60

1 2

3 4

Solution:

The solution is simple when the cylinder configuration of the engine is known.

Figure S2.7a shows how the cylinders are arranged. According to this configuration,

cylinders 1 and 4 are both at TDC while the two others are at BDC. Thus this engine

works exactly as a 4-cylinder in-line engine works. One possibility is to consider the

cylinder 2 at compression when the cylinder 1 is at ignition. This results in 1-2-4-3

firing order as depicted in Figure S2.7b. A 1-3-4-2 firing order is also possible by

changing the stroke orders of cylinders 2 and 3.

Figure S2.7a

Figure S2.7b Firing order chart for Problem 2.7

Problem 2.8

Construct the firing map for a 6 cylinder in-line engine with cranks at 0-240-120-0-240-120

degrees.

1

2

3

Relativ

e state of stro

kes

Exhaust

Cran

k lay

ou

t

Stroke order

Firing order

0

60

180 180 180

1, 2

3, 4

Compression


Intake

180


1 2 4 3

4


Solution:

Comparing this layout with that of Figure 2.55 reveals that only the conditions of

cylinders 4 and 6 are changed. Therefore, the construction of the firing map is quite

similar to that of Figure 2.55. The result is shown in Figure S2.8.

1

2

3

4

1 4 3 2

Relativ

e state of stro

kes

Exhaust

Exhaust

Exhaust

Exhaust

Intake

Intake

Intake

Intake

Compression

Compression

Compression

Compression

Cran

k lay

ou

t

Stroke order

Firing order

0

180

180

0

180 180 180 180

Solution:

a) From the discussions of Section 2.4.1 for the firing order of an in-line 4 cylinder

engine with crank layout of 0-180-180-0, it may be recalled that two firing orders of

1-3-4-2 and 1-2-4-3 are possible. For the firing order of 1-4-3-2, therefore, the crank

layout must be different. The obvious difference between 1-4-3-2 and 1-3-4-2 firing

orders is the exchange of cylinders 3 and 4. For this reason the 0-180-180-0

configuration can be changed to 0-180-0-180. The firing map for this layout is

shown in Figure S2.9.

b) According to Equation 2.115, the crank angles i are 1, 2=1-, 3=1-2 and

4=1-3. The state angle Si shows the state of a cylinder relative to that of cylinder

1 that is at ignition. According to the firing diagram of Figure S2.9, cylinders 2 to 4

are at exhaust, intake and compression respectively. Thus the state angles are

01 S , 2S , 23 S and 34 S for cylinders 1-4.

Figure S2.9 Firing map for Problem 2.9

Problem 2.10

Compare the torque outputs of an inline four cylinder, four stroke engine at two firing orders of

1-3-4-2 and 1-2-4-3 using the information of Example 2.4.3.

Problem 2.11

Use the information of Example 2.4.3 to plot the variation of the torque of the three cylinder

engine of Problem 2.6 and calculate the flywheel inertia.

Solution:

The solution for 1-3-4-2 firing order was performed in Example 2.4.3 with

MATLAB program listing of Figure 2.60. In order to solve for the 1-2-4-3 firing

order, only the state angles must be changed. These are 01 S , 32 S , 3S

and 24 S for cylinders 1-4. The only change in the program is:

DF=[0 3*pi pi 2*pi]; % State angle of cylinders

The result for the torque output of engine as would be expected, will be exactly

similar to Figure 2.61 belonging to 1-3-4-2 firing order.

0 90 180 270 360 450 540 630 720 -150

-100

-50

0

50

100

150

200

First cylinder crank angle (deg)

To

tal engin

e t

orq

ue (

Nm

)

(Nm

) A1=-6

A2=130

A3=-120

A4=2

A5=-12

A6=130

Solution:

The solution is quite similar to Problem 2.10 as the same MATLAB program can be

used. The changes in the program are 1) number of cylinders and 2) state angle

inputs. The state angles for this engine must be obtained according to Figure S2.6.

The state of cylinder 2 is 60 degrees before the compression state, or 120 degrees

after the intake state. This means 3

83

22

332

S . The state of

cylinder 3 similarly is 3

43

223

S . Therefore in the MATALB program:

DF=[0 8*pi/3 4*pi/3]; % State angle of cylinders

The result for this engine is shown in Figure S2.11.

In order to calculate the flywheel inertia, the areas under the engine torque curve

must be measured. This is a difficult process and only approximate values are

obtained here as shown in Table S2.11. It is concluded that min and Max (see

Section 2.3.4) occur before and after each large positive area. Therefore the area A*

equals (see Equation 2.114) 124 - (- 6) =130 (Joule) and for a fluctuation index of

0.02, the flywheel inertia can be calculated from Equation 2.113:

2

207.0

)30/3000(02.0

130kgmIe

Figure S2.11 Torque variation of the 3-cylider engine of Problem 3.11

Table S2.11 Areas under the engine torque curve of Problem 2.11

Item 1 2 3 4 5 6

Ai (Joule) -6 130 -120 2 -12 130

A (Joule) -6 124 4 6 -6 124

min A Amin Amin

Max A AMax AMax

Problem 2.12

Compare the variation of the torque of the 4 cylinder V engine of Problem 2.7 with an in-line

layout. Use the information of Example 2.3.3.

Problem 2.13

Plot the variations of engine power losses with altitude and temperature changes for Example

2.8.2 in SI units.

Solution:

According to the solution of Problem 2.7, the firing map of the V engine is exactly

similar to that of the in-line engine (see Figure S2.7). Therefore, the torque

variations of the both engines are similar. Also, based on the result of Problem 2.10,

the torque variations are the same as that for a 1-3-4-2 firing order given in Figure

2.61.

Figure S2.13a Engine power loss with altitude

0 5 10 15 20 250

500

1000

1500

2000

2500

3000

Power loss (%)

Alti

tud

e (

m)

Solution:

The first part can be obtained from available program listing by simply removing

*3.2808 term from the plot statement. The result is shown in Figure S2.13a.

The second part of solution can be obtained by using the following simple program:

T=T0: 0.2: 323.16; % Up to 50 deg C

cft=sqrt(T0)./T.^0.5;

p_loss=100*(1-cft);

figure

plot(T-273.16, p_loss)

grid

xlabel('Temperature (C)')

ylabel('Power loss (%)')

The result is depicted in Figure S2.13b.

Figure S2.13b Engine power loss with temperature increase

Problem 2.14

The variation of gas pressure of a single cylinder 4 stroke engine during 2 complete revolutions

in speed of 2000 rpm is simplified to the form shown in Figure p.2.14. Other engine parameters

are given in Table p.2.14.

Figure p.2.14 Cylinder pressure

Table p.2.14 Engine parameters

Parameter value

Cylinder diameter 10 cm

Crank radius 10 cm

Connecting rod cent-cent 25 cm

20 25 30 35 40 45 500

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

Temperature (C)

Pow

er

loss

(%)

0

0.5

1

1.5

2

2.5

3

3.5

0 1 2 3 4

Pre

ssu

re (

Mp

a)

Engine revolution (x)

Con rod CG to crank axis 7 cm

Con rod mass 1.0 kg

Piston mass 1.0 kg

Use the simplified engine model and,

a) Find the equivalent mass mA for the connecting rod. b) Calculate the inertia force FIP in terms of crank angle and engine speed. c) Write an equation for the total vertical force FBy acting at point A. d) Plot the variation of FBy and FW versus crank angle. e) Plot the variation of torque versus crank angle. f) Find the average engine torque and compare it with the quasi-steady torque resulting from

average pressure during combustion phase.

g) Determine the mean effective pressure for the engine.

Solution:

a) According to Equation 2.96:

kgml

lm BA 28.01

25

7

b) From Equation 2.102 for FIP and Equations 2.42 and 2.50 for the piston

acceleration:

22 )2cos4.0(cos128.0)2cos(cos)( eeAPIP Rl

RmmF

c) According to Equation 2.105:

IPPBy FFF

where FP is the pressure force P.AP. Therefore:

23 )2cos4.0(cos128.0)(1085.7)( eIPPBy PFAPF

d) At 2000 rpm P() is given in Figure P2.14 and FBy can be determined at each

crank angle . FW according to Equation 2.92 is also dependent on FBy:

tanByW FF

in which is given by Equation 2.48:

sin4.0sin 1

The pressure distribution can be divided into 4 zones:

43;)3(101

3;04

;)1(104

40;)81(101

6

6

6

P

Solution (continued):

FBy therefore can be written as:

43;)2cos4.0(cos5615250023562

3;)2cos4.0(cos56154

;)2cos4.0(cos56151000031416

40;)2cos4.0(cos5615200007854

ByF

Similarly the equations for the variations of FW can be written. The variations of FBy

and FBx (FW) with crank angle are plotted in Figure S2.14a.

e) Based on Equation 2.103, the engine torque simply is hFT We . However, h has a

complicated relation with . From Equation 2.109:

sin4.0sincos25.0cos1.0 1h

Therefore:

sin4.0sintansin4.0sincos25.0cos1.0 11 Bye FT

And the variation of torque with the crank angle will be that shown if Figure S2.14b.

f) The average torque is 181 Nm. The average pressure during the combustion phase

is:

8

13

8

13

avP MPa

The torque resulting from the average torque can be obtained by replacing pbme with

Pav in Equation 2.124:

ave

av Ps

VT

in which

333 0024.010)250100(10854.7)( mlRAV Pe

NmPs

VT av

eav 30510

8

13

4

0024.0 6

Figure S2.14a Piston forces

0 90 180 270 360 450 540 630 720-1

-0.5

0

0.5

1

1.5

2x 10

4

Crank angle (deg)

Pis

ton

forc

es (

N)

FBy

FBx


g) The mean effective pressure according to Equation 2.124 is:

MPaTV

sp av

e

bme 965.01810024.0

4

Figure S2.14b Engine Torque

Problem 2.15

The torque-angle relation for a four cylinder engine at an idle speed of 1000 rpm is of the form:

)(2cos 00 aTTT

a) Find the area A* of torque fluctuations relative to the average engine torque.

b) Show that the value of necessary flywheel inertia can be written as akTI .

c) For a value of 2% permissible speed fluctuations, evaluate k.

Results: (a) Ta , (c) 0.0046

0 90 180 270 360 450 540 630 720-700

0

700

1400

2100

Crank angle (deg)

Eng

ine

to

rqu

e (

Nm

)

Figure S2.15 Engine torque shape

0 0.5 1 1.5 2 2.5 3 3.5 4

0

T0

Engin

e t

orq

ue

First cylinder crank angle ()

/2

/2

A*

m

M

Solution:

a) The engine torque curve is depicted in Figure S2.15. Each of the equal areas in the

figure relative to the average torque T0 is the area A* that is:

M

m

M

m

dTdTTA a

)(2cos)( 00

*

m and M are the points at which T=T0, that is:

0)(2cos 0

with solutions: ...,4

3,4

0

Considering 04

3

m and 04

5

M , the result of integrating for A* is:

aa TdTA

0

0

45

43

0

* )(2cos

b) According to Equation 2.113 the flywheel inertia is:

a

avFavF

e Tii

AI

22

* 1

which is in the form of akTI .

c) k simply is:

0046.0

)30

1000(02.0

11

22

avFi

k

Figure S2.15 Engine torque shape

0 0.5 1 1.5 2 2.5 3 3.5 4

0

T0

Engin

e t

orq

ue

First cylinder crank angle ()

/2

/2

A*

m

M

Problem 3.1

The rolling resistance force is reduced on a slope by a cosine factor ( cos ). On the other hand,

on a slope the gravitational force is added to the resistive forces. Assume a constant rolling

resistance force and write the parametric forms of the total resistive force for both cases of level

and sloping roads. At a given speed v0,

a) Write an expression that ensures an equal resistive force for both cases.

b) Solve the expression obtained in (a) for the parametric values of corresponding slopes.

c) For the coefficient of rolling resistance equal to 0.02, evaluate the values of the slopes

obtained in (b) and discuss the result.

Result: (a) fR (1- cos ) = sin

Solution:

a) It is required that the total resistive force on a level road to be equal to that of a

sloping road at a certain forward speed v0. In mathematical form:

2

0

2

0 sincos cvWWfcvWf RR

or,

sincos RR ff

which is an expression that relates the relevant parameters.

b) Solution of the above expression found in part (a) can be obtained by writing the

trigonometric functions in half-arc forms:

0)2

cos2

sin(2

sin

Rf

which has the two following answers:

R

Rf

f1

2tan0

2cos

2sin

002

sin

c) Apart from the trivial solution =0, the second solution for fR=0.02 results in

=177.7! (Note that even for fR=1 the slope is 90). Thus it is concluded that there is

no slope for which the total resistive force at a given speed will be equal to that on

the level road at the same speed.

Problem 3.2

For the vehicle of Example 3.4.2,

a) Calculate the overall aerodynamic coefficient for the same temperature at altitude of 1000 m.

b) Repeat (a) for the same altitude at temperature 30 C.

c) At the same altitude of (a) at what temperature the drag force increases by 20%?

d) At the same temperature of (a) at what altitude the drag force reduces by 20%?

Solution:

a) From Equation 2.138, the air density for the specified altitude can be obtained:

143.1225.110001021.911021.91 714.050714.05 H

therefore the overall aerodynamic coefficient is:

435.00.238.0143.15.0 c

b) At the same altitude the air pressure is (Equation 2.137):

92005.10132)1021.91()1021.91( 205 pHp

and from Equation 3.30:

056.13015.273

92000348.00348.0

T

PA

the overall aerodynamic coefficient therefore is:

401.00.238.01056.15.0 c

c) In order to have a 20% increase in the drag force the air density must increase by

20% (i.e. =1.372). Therefore:

CKP

TA

8.394.233372.1

92000348.00348.0

d) The air density at (a) is 1.225. In order to have a decrease of 20% in the

aerodynamic force, the air density must decrease by the same amount. Thus from

Equation 2.138:

714.051021.918.0 H

which gives H = 2914 m.

Problem 3.3

In order to have a rough estimation for the performance of a vehicle, it is proposed to ignore the

resistive forces to obtain the No-Resistive-Force (NRF) performance.

a) Derive the governing equations of vehicle longitudinal motion for speed v(t) and distance S(t)

by neglecting all resistive forces for the CPP (see Section 3.5).

b) For a vehicle of mass 1.2 ton, determine the required engine power P for achieving

acceleration performance of 0-100 km/h during 10, 8 or 6 seconds.

c) Evaluate the power increase factors from 10 seconds to t* seconds defined as

[P/P10=[P(t*)-P(10)]/P(10)], for t

*=8 and 6.

Results: (a) m

Ptvv

220 ,

P

vvmS

3

)( 303

, (b) 46.3, 57.9 and 77.2 kW, (c) 0.25 and 0.67

Solution:

a) By neglecting all resistive forces in Equation 3.58 the equation of motion is:

v

P

dt

dvm

Integrating with initial condition of v=v0 @ t=0; results in:

m

Ptvv

220

Using the equation relating speed to acceleration and distance, namely:

adSvdv

and substituting in the first equation leads to:

PdSdvmv 2

Integrating with the initial condition of S=0 @ t=0; results in:

P

vvmS

3

)( 303

b) Since the vehicle starts acceleration from rest (v0=0), therefore:

tP

2

)6.3/100(1200 2

Results for t=10, 8 and 6 seconds are 46,300, 57,870 and 77,160 W respectively.

c) The increased power factor is:

)10(

)10()(/

*

10P

PtPPP

For t*=8 and 6 second, the results are 0.25 and 0.67.

Problem 3.4

Use the results of Problem 3.3 and,

a) Write the expression for the specific power Ps (in W/kg) of a vehicle to reach a certain speed v

(km/h) from the rest at a certain acceleration time t.

b) Plot the variation of Ps versus t from 6 to 10 seconds. Repeat the result for three speeds of 80,

90 and 100 km/h.

c) Are the results dependent on the vehicle properties?

Solution:

a) By assuming v=0 @ t=0; the speed relation reduces to:

m

Ptv

2

The specific power Ps defined as P/m, therefore, becomes:

t

vPs

2

2

To obtain Ps in W with speed in km/h:

t

vPs

92.25

2

b) For the three speeds of 80, 90 and 100 km/h, the plots are given in Figure S3.4.

c) No, the results are valid for all vehicles as long as no resistive forces are present.

Thus this figure can be used to estimate the necessary power just by multiplying the

specific power by the vehicle mass.

Figure S3.4 Specific power requirements at no resistive force

Problem 3.5

At very low speeds the aerodynamic force is small and may be neglected. For example, at speeds

below 30 km/h, the aerodynamic force is one order of magnitude smaller than the rolling

resistance force. For such cases categorised as Low-Speed (LS), ignore the aerodynamic force

and for the CPP assume a constant rolling resistance force F0, and,

a) Integrate the equation of motion (Equation 3.58 with c=0) and use the initial condition of v=v0

at t=t0 to obtain an expression for the travel time in terms of speed.

b) For a vehicle of 1000 kg mass and total rolling resistance force of 200 N, when starting to

move from standstill, plot the variation of vehicle speed against elapsed time up to 10 seconds

and compare it with the results of NRF model (Problem 3.3). The engine power is 50 kW.

Results: (a) vFP

vFP

F

Pm

F

vvmtt

0

00

2

00

00 ln

)(

6 6.5 7 7.5 8 8.5 9 9.5 1020

25

30

35

40

45

50

55

60

65

Time of travel (s)

Spe

cific

pow

er

(W/k

g)

80 km/h

90 km/h

100 km/h

Solution:

a) For such cases the differential equation of motion will read:

0Fv

P

mdvdt

This can be integrated to obtain:

101 )ln( mvvFPmpCt

where, C is the constant of integration, kWF 0 , 20

1F

Pp and

0

1F

vv . For a general

initial condition of t=t0 and v=v0 the result is:

0

0

0

0010

)(ln

F

vvm

vFP

vFPmptt

b) For motions starting from rest, the equation is:

1

11

11 ln mv

vp

pmpt

In order to obtain the variation of speed with time, the values of speed ranging from

0 up to 30 m/s are given and the corresponding time values are obtained. For the

NRF model the speed relation was earlier found to be m

Ptv

2 .

The variation of vehicle speed with time for P=50,000 W is shown in Figure S3.5. It

is clear that up to speeds of 40-50 km/h the differences are very small.

Figure S3.5 The time history of vehicle speed

Problem 3.6

For the vehicle of Problem 3.5 and using the LS method, find the required power for the 0-100

km/h acceleration to take place in 7 seconds.

Result: 58,849 W.

Hint: The following statements in MATLAB can be used with a proper initial guess for x0.

fun=inline(7-1000*x*log(x/(x-(100/3.6/200)))+1000*(100/3.6/200));

x=fsolve(fun, x0, optimset('Display','off')); (x = P/F0^2)

0 2 4 6 8 100

5

10

15

20

25

30

Time (s)

Velo

city

(m/s

)

LS model

NRF model

Problem 3.7

For the vehicle of Problem 3.5 using the LS method determine the power requirements for a

performance starting from rest to reach speed v at time t, for 3 cases of v=80, 90 and 100 km/h

for accelerating times varying from 6 to 10 seconds. Plot the results in a single figure.

Solution:

In MATLABs command window simply type the first statement given:

fun=inline('7-1000*x*log(x/(x-(100/3.6/200)))+1000*(100/3.6/200)');

then specify an initial value for x (i.e. x0). Any value between 1 and 10 would be

fine. Then type the second statement:

x=fsolve(fun, x0, optimset('Display','off'))

the result will be x=1.4712. x was defined as x = P/F0^2, so:

P = 1.47122002 = 58,849 W.

Figure S3.7 Power requirements for different acceleration times

6 6.5 7 7.5 8 8.5 9 9.5 1025

30

35

40

45

50

55

60

65

70

Time (s) to reach a certain speed v (km/h)

Pow

er

req

uir

ed

(kW

)

v=80

v=90

v=100

Solution:

In Problem 3.6 a single point was considered. Here several similar calculations are

needed. A loop, therefore, can be created to repeat the calculations for all the points

in the range. A sample program in MATLAB is given below:

t=6: 0.02: 10; % Time span

x0=1; % Initial condition

for j=1:3 % Loop for speed

v=(80+(j-1)*10);

for i=1: length(t) % Loop for time

P0=fsolve(@(x) t(i)-1000*x*log(x/(x-(v/3.6/200)))+1000*(v/3.6/200), x0, ...

optimset('Display','off'));

P(i)=P0*F0^2;

x0=P0; % Initial condition for the next iteration

end

plot(t, P/1000)

hold on

end

xlabel('Time (s) to reach a certain speed v (km/h)')

ylabel('Power reqired (kW)')

grid

legend('v=80', 'v=90', 'v=100')

The result is shown in Figure S3.7.

Problem 3.8

The power evaluation for the NRF case (Problem 3.3) is a simple closed-form solution but it is

not accurate. The LS method (problems 3.5-3.7) produces more accurate results especially in the

low speed ranges. By generating plots similar to those of Problem 3.7 show that an approximate

equation of P=PNRF+0.75F0v can generate results very close to those of LS method.

Solution:

At the end of the MATLAB program of Problem 3.7 the simple calculations of NRF

model can be included and then the approximate model can be evaluated. The

following sample listing may be used:

for j=1: 3

v=80+(j-1)*10;

Ps=(m*v^2/25.920)./t; % NRF model

PA=(Ps+0.75*F0*v/3.6)/1000; % Approximate model

plot(t, PA, '-.')

end

The plots of approximate model are included in the same figure obtained in Problem

3.7. The result is Figure S3.8. The results of the two methods are very close.

Figure S3.8 Comparison between the LS and approximate models

Problem 3.9

For the LS case use adSvdv that relates the speed to acceleration and distance, substitute for

acceleration in terms of speed and,

a) Integrate to obtain an expression for travel distance S in terms of velocity v.

b) Derive the equation for a motion starting at a distance S0 from origin with velocity v0.

c) Simplify the expression for a motion stating from rest at origin.

Results: (a) ]5.0)ln([ 112

10

2

10 vpvvFPpmFCS , (c) )5.0ln( 112

1

11

12

10 vpvvp

ppmFS

With 2

0

1F

Pp and

0

1F

vv .

6 6.5 7 7.5 8 8.5 9 9.5 1025

30

35

40

45

50

55

60

65

70


Pow

er

req

uir

ed

(kW

)

v=80

v=90

v=100

Approximate

Solution:

a) Acceleration a is in the form of:

)(1

0Fv

P

ma

Travel distance can be obtained by the application of the given relation:

dSFv

P

mvdv )(

10

or,

vFP

dvmvdS

0

2

Integration by separation of variables will lead to:

]5.0)ln([ 112

10

2

10 vpvvFPpmFCS

where C is the constant of integration.

b) For a general case of starting the motion at a distance S0 from origin with velocity

v0,

]5.0)ln([0

012

0

2

000

2

100F

vp

F

vvFPpmFSC

and

)()(5.0ln 1

0

01

2

12

0

2

0

0

002

100 vF

vpv

F

v

vFP

vFPpmFSS

c) For a motion stating from rest at origin (S0=0, v0=0:

)5.0ln( 112

1

11

12

10 vpvvp

ppmFS

Problem 3.10

A vehicle of 1200 kg mass starts to accelerate from the rest at origin. If power is constant at 60

kW, for a LS model with F0=200 N, determine the travel time and distance when speed is 100

km/h. Compare your results with those of NRF model.

Results: t= 8.23 s, S=153.65 m for LS and t= 7.72 s and S=142.9 m for NRF.

Problem 3.11

In Problem 3.8 a close approximation was used for the power estimation of LS method. For the

general case including the aerodynamic force, the approximation given by P = PNRF+0.5FRv

is found to work well.

Solution:

From Problems 3.5 and 3.9 for motion starting from rest at origin (S0=0, v0=0):

1

11

11 ln mv

vp

pmpt

)5.0ln( 112

1

11

12

10 vpvvp

ppmFS

with 2

0

1F

Pp and

0

1F

vv . Using the information for the current problem:

50.1200

6000022

0

1 F

Pp and 1389.0

2006.3

100

0

1

F

vv

Therefore:

st 23.81389.012001389.05.1

5.1ln5.11200

mS 65.153)1389.05.11389.05.01389.05.1

5.1ln5.1(2001200 22

For the NRF model the same quantities can be determined from relations obtained in

Problem 3.3 (S0=0, v0=0):

sP

vmt 72.7

600006.32

1001200

2 2

22

mP

mvS 9.142

600006.33

1001200

3 3

33

For the vehicle of Example 3.5.3 plot the variations of power versus acceleration times similar to

those of Problem 3.8 and compare the exact solutions with those obtained from the proposed

method.

Solution:

Here the exact solution is first needed. For this purpose the MATLAB program of

Example 3.5.3 can be modified in order to repeat the power calculations at several

desired points. The following program can be used:

t=6: 0.02: 10; % Time span

for j=1: 3 % Loop for speed

vd=(80+(j-1)*10)/3.6;

Ps=(F0+c*vd^2)*vd+1; % Initial guess for power

for i=1: length(t) % Loop for time

td=t(i);

P0=fsolve(@f_353, Ps, optimset('Display','off'));

P(i)=P0;

Ps=P0; % Initial condition for the next iteration

end

plot(t, P/1000)

hold on

end

At the next step, the approximation given by P = PNRF+0.5FRv should be

constructed. Note that instead of F0 in Problem 3.8, this time FR=F0+cv2 must be

used in the approximation equation.

The results plotted in Figure S3.11 show good agreement. Therefore, the power

estimation for an acceleration performance to a desired speed vd at a desired time td

can be performed accurately by the following simple formula:

dd

d

d vcvFt

mvP )(

2

1

2

2

0

2

Figure S3.8 Comparison between the exact and approximate models

Problem 3.12

According to the solutions obtained for CTP (see Section 3.6) it turned out that at each gear, the

acceleration is constant to a good degree of approximation (see Figure 3.50). Thus a simpler

solution can be obtained by considering an effective resistive force for each gear that reduces the

problem to a Constant Acceleration Approximation (CAA). In each gear assume the resistive

force acting on the vehicle is the average of that force at both ends of the constant torque range.

Write the expressions for the average speed at each gear vav, the average resistive force Rav and,

a) Show that the acceleration, velocity and distance at each gear are

)(1

avTii RFm

a ,

ivttatv ii 00 )()( and ii Oii SttvttaS )()(5.0 00

2

0 , in which )1(max0 ivv i and

)1(max0 iSS i are the initial speed and distance from origin for each gear for i>1 and v0 and S0

for i=1.

b) Repeat Example 3.6.1 by applying the CAA method.

6 6.5 7 7.5 8 8.5 9 9.5 1025

30

35

40

45

50

55

60

65

70

75


Pow

er

req

uir

ed

(kW

)

v=80

v=90

v=100

Approximate

Solution:

a) The average speed at each gear is (no slip condition is assumed):

av

i

wav

n

riv )(

where:

)(5.0 21 av

and the average resistive force, therefore, is:

2

0 avav cvFF

Then the acceleration at each gear is:

cteFFm

a avTii )(1

As the acceleration is constant, velocity and travel distance at each gear are:

ivttatv ii 00 )()(

iiSttvttaS ii 000

2

0 )()(5.0 (15)

In which i

v0 and S0i are the initial speed and distance from origin at the start of

motion in each gear. For the first gear 10

v is the initial speed of vehicle (often zero),

and at other gears it is the maximum attained speed before the gearshift:

)1(max0 ivv i , i>1

Similarly the final distance travelled in previous gear is the initial distance for the

next gear:

)1(max0 iSS i , i>1


b) The average engine speed is 2000 rpm. The calculated results for average speed at

each gear, accelerations, final speeds, travel times and distances are given in Table

S3.12. The final speed at each gear is calculated from:

M

i

w

n

riv )(max

The final travel time at each gear is:

i

ia

vivtt i

0max

0max

)(

At each tmax, the final travel distance can be determined.

To plot the variation of parameters, the time span is divided into several small time

intervals and then each parameter is evaluated at each time value. The results for the

acceleration, speed and distance are shown in Figures S3.12a - S3.12c. A MATLAB

program to generate the results is given below:

m=2000; fR=0.02; Ca=0.5; rW=0.3; nf=4.0; Tm=220; % Constant torque (Nm)

wm=1200; % Minimum engine speed (rpm)

wM=2800; % Maximum engine speed (rpm)

n_g=[5.0 3.15 1.985 1.25]; % Transmission ratios 1-4

n=n_g*nf; % Total gear ratios

F0=m*9.81*fR;

t0=0; v0=0; s0=0; % Initial conditions

w_av=0.5*(wm+wM)*pi/30;

t0i=t0; v0i=v0; s0i=s0;

for i=1: length(n_g)

vav(i)=rW*w_av/n(i);

FT(i)=n(i)*Tm/rW;

Fav(i)=F0+Ca*vav(i)^2;

a(i)=(FT(i)-Fav(i))/m;

vmax(i)=wM*rW*pi/n(i)/30;

tmax(i)=t0i+(vmax(i)-v0i)/a(i);

smax(i)=s0i+0.5*a(i)*(tmax(i)-t0i)^2+v0i*(tmax(i)-t0i);

t=t0i: 0.02: tmax(i);

acc=a(i)*ones(1, length(t));

v=v0i+a(i)*(t-t0i);

s=s0i+0.5*a(i)*(t-t0i).^2+v0i*(t-t0i);

if i>1 acc(1)=a(i-1); end

figure(1), plot(t, acc), hold on

figure(2), plot(t, v), hold on figure(3), plot(t, s), hold on

t0i=tmax(i);

v0i=vmax(i);

s0i=smax(i);

end

Table S3.12 Results for Problem 3.12

vav

(m/s)

Fav

(N)

FT

(N)

a

(m/s^2)

vmax

(m/s)

ti

(s)

Si

(m)

Gear 1 3.142 397.34 14667 7.135 4.398 0.617 1.356

Gear 2 4.987 404.83 9240 4.418 6.981 1.201 3.327

Gear 3 7.915 423.73 5821 2.699 11.079 2.719 13.706

Gear 4 12.566 471.36 3667 1.598 17.593 6.796 58.453

Figure S3.12a Variation of acceleration

Figure S3.12b Variation of speed

0 1 2 3 4 5 6 71

2

3

4

5

6

7

8

Time (s)

Acc

ele

ration

(m

/s2

)

0 1 2 3 4 5 6 70

2

4

6

8

10

12

14

16

18

Time (s)

Spe

ed

(m

/s)

Figure S3.12c Variation of distance

Problem 3.13

A 5th

overdrive gear with overall ratio of 3.15 is considered for the vehicle in Problem 3.12, and

the torque is extended to 3400 rpm. Obtain the time variations of acceleration, velocity and travel

distance for the vehicle by both CAA and numerical methods and plot the results.

0 1 2 3 4 5 6 70

10

20

30

40

50

60

70

80

Time (s)

Dis

tan

ce (

m)

Solution:

The CAA solution of this problem is obtained just by inclusion of the fifth gear and

changing the maximum engine speed in the MATLAB program of Problem 3.12:

wM=3400; % Maximum engine speed (rpm)

n_g=[5.0 3.15 1.985 1.25 3.15/nf]; % Transmission ratios 1-5

Solution with the numerical method was examined in Example 3.6.2 with MATLAB

program of Figure 3.53. The same program with small modifications can be used

here as well. Results for the speed and distance are compared in Figures S3.13a and

S3.13b. The dashed plots belong to the numerical method.

Figure S3.13a Comparison between speeds of CAA (solid) and numerical (dashed) methods

Figure S3.13b Comparison between distances of CAA (solid) and numerical (dashed) methods

0 5 10 15 20 250

5

10

15

20

25

30

35

Time (s)

Spe

ed

(m

/s)

0 5 10 15 20 250

100

200

300

400

500

600

Time (s)

Dis

tan

ce (

m)

Problem 3.14

In Example 3.5.2 impose a limit for the traction force of FT < 0.5 W and compare the results.

Figure S3.14 Comparison of vehicle speed with and without traction limit

0 10 20 30 40 50 60 70 800

5

10

15

20

25

30

35

40

45

50

Time (s)

Velo

city

(m/s

)

With limit

No limit

Solution:

By defining a road adhesion coefficient R in the MATLAB program of Example

3.5.2 and including it in the global statement,

global p c f0 m Ftmax

mio_r=0.5; % Road adhesion coefficient

Ftmax=mio_r*m*9.81;

Then the limit for the tractive force can be included in the function const_pow:

% Impose a limit on the traction force

if ft > Ftmax, ft=Ftmax; end

f=(ft-f0-c*v^2)/m;

The results with and without traction limit are plotted in Figure S3.14. The reason

for the small difference is that the tyre traction force is usually smaller than the

adhesion limit during the vehicle motion. Only at low speeds does the tyre traction

become large.

Problem 3.15

For a vehicle with transmission and engine information given in Example 3.7.2, include a one-

second torque interruption for each shift and plot similar results. To this end, include a

subprogram with listing given below at the end of loop for each gear:

% Inner loop for shifting delay:

if i

Figure S3.15a Speed and distance outputs of Problem 3.15

Figure S3.15b Acceleration output of Problem 3.15

0 10 20 30 40 50 600

10

20

30

40

50

Velo

city

(m/s

)

0 10 20 30 40 50 600

500

1000

1500

2000

2500

Time (s)

Dis

tan

ce (

m)

0 10 20 30 40 50 60-1

0

1

2

3

4

5

6

7

Time (s)

Acc

ele

ration

(m

/s2

)

Problem 3.16

Repeat Problem 3.15 with a different shifting delay for each gear of the form 1.5, 1.25, 1.0 and

0.75 second for 1-2, 2-3, 3-4 and 4-5 shifts respectively. (For this you will need to change the

program).

0 10 20 30 40 50 600

10

20

30

40

50

Velo

city

(m/s

)

0 10 20 30 40 50 600

500

1000

1500

2000

2500

Time (s)

Dis

tan

ce (

m)

Solution:

Including different shifting delays in the MATLAB program of Problem 3.15 is a

very simple task. Instead of defining tdelay=1; for the shift delay time in the input

section of the program, tdelay is defined as an array:

tdelay=[1.5 1.25 1.0 0.75];

In addition the statement tf=t0+tdelay; is modified to:

tf=t0+tdelay(i);

No other change is necessary and the results will be similar to those given in Figures

S3.16a and S3.16b.



Problem 3.17

Repeat Example 3.7.2 for a different shifting rpm.

a) Shift all gears at times when the engine speed is 4500 rpm.

b) Shift the gears at 4500, 4000, 3500 and 3000 rpm for shifting 1-2, 2-3, 3-4 and 4-5

respectively. (For this part you will need to change the program).

0 10 20 30 40 50 60-1

0

1

2

3

4

5

6

7

Time (s)

Acc

ele

ration

(m

/s2

)

Solution:

Running the MATLAB program of Example 3.7.2 for a new shift rpm given in part

(a) is straightforward. However, in order to make the program general, instead of

defining a single value wem for the shift rpm of all gears, w_shift is defined as

an array that contains the shift rpms of all gears:

w_shift=[w1 w2 . wn];

where w1, w2, etc are the shift rpms for gear 1, gear 2 and so on. Note that wn

means the shift rpm for the final gear and is meaningless as no (up)shift takes place

for the final gear. So a large number can be considered for it (e.g. 8000 rpm). But at

the same time all we=6500; statements in the program also must be increased to a

value larger than that (e.g. we=max(w_shift)+10 rpm).

In addition a new statement wem=w_shift(i); needs to be included in the beginning of

the loop for gears.

a) Just insert w_shift=[4500 4500 4500 4500 8000]; and run the program. The

results are shown in Figures S3.17a and S3.17b.

b) This time insert w_shift=[4500 4000 3500 3000 8000];.

The results will be similar to those given in Figures S3.17c and S3.17d.

Figure S3.17a Speed and distance outputs of Problem 3.17, part (a)

Figure S3.17b Acceleration output of Problem 3.17, part (a)

0 10 20 30 40 50 600

10

20

30

40

50

Velo

city

(m/s

)

0 10 20 30 40 50 600

500

1000

1500

2000

2500

Time (s)

Dis

tan

ce (

m)

0 10 20 30 40 50 600

1

2

3

4

5

6

7

Time (s)

Acc

ele

ration

(m

/s2

)

Figure S3.17c Speed and distance outputs of Problem 3.17, part (b)

Figure S3.17d Acceleration output of Problem 3.17, part (b)

Problem 3.18

In Example 3.7.3, investigate the possibility of having a dynamic balance point at gear 4. In case

no steady state point is available, find a new gear ratio to achieve a steady-state.

0 10 20 30 40 50 600

10

20

30

40

50

Velo

city

(m/s

)

0 10 20 30 40 50 600

500

1000

1500

2000

Time (s)

Dis

tan

ce (

m)

0 10 20 30 40 50 600

1

2

3

4

5

6

7

Time (s)

Acc

ele

ration

(m

/s2

)

Problem 3.19

Repeat Example 3.7.2 with transmission ratios 3.25, 1.772, 1.194, 0.926 and 0.711.

Solution:

For gear 4 the overall ratio is n4=1.14=4.4 and k1 and k2 are:

167.3810058.4)27.0/4.4(35.0

3027.0)27.0/4.4(

)/(

)/(43

2

1

3

2

2

1

trnc

trnk

wi

wi

70427.0

24.554.481.9100002.0302

w

ir

tnFk

The solution of 2*)(v +k1*v +k2=0 with MATLAB is: vstar=roots([1 k1 k2]). The

answers are v*=51.767 and -13.60. The engine speed at the positive answer is

*=8,056 rpm that is unacceptable. Thus, there is no dynamic balance point for the

gear 4.

If a maximum engine speed of 6000 rpm is considered, then the maximum speed

should be v*=6000rw/30/n4. The problem now is to find the unknown n4/rw in

the equation 2*)(v +k1*v +k2=0 so that v

* equals (6000/30)/(n4/rw). This can be

solved by using fsolve function in the statement below:

NRW=fsolve(@(x) (6000*pi/30/x)^2-(6000*pi/30/x)*(x^2*0.3027/(c+x^3*4.058e-4))...

+196.2-x*55.24, 10, optimset('Display','off'))

x in the equation stands for the ratio n4/rw, 196.2 is the value of F0 and 10 is an

initial guess for x. The answer for the above statement is NRW=12.332. Then n4

is: n4= NRWrw/nf = 0.8324.


0 10 20 30 40 50 600

10

20

30

40

50

Velo

city

(m/s

)

0 10 20 30 40 50 600

500

1000

1500

2000

2500

Time (s)

Dis

tan

ce (

m)

Solution:

Inclusion of the new gear ratios in the MATLAB program of Example 3.7.2 is a very

simple task and the results will look like those given in Figures S3.19a and S3.19b.

Comparing the results with those of Example 3.7.2 shows that with the new gear

ratios the shift times are changed considerably but the final performance at 60

seconds is almost similar with slight improvements for the new gear set. The

maximum speed is increased from 48 to 48.24 m/s (0.5%) and travel distance from

2267 to 2282 m (0.7%). The acceleration time to 100 km/h, however, is slightly

longer increasing from 10 to 10.5 seconds (5%).


Problem 3.20

In the program listing given for Example 3.7.2 no constraint is imposed for the lower limit of

engine speed and at low vehicle speeds the engine rpm will attain values less than its working

range of 1000 rpm.

a) For the existing program try to find out at what times and vehicle speeds the engine speed is

below 1000 rpm.

b) Modify the program to ensure a speed of at least 1000 rpm for the engine. How are the results

affected?

0 10 20 30 40 50 600

1

2

3

4

5

6

Time (s)

Acc

ele

ration

(m

/s2

)

Solution:

a) Since the vehicle speed and engine speed arrays are already available in the

MATLAB program of Example 3.7.2, the variation of engine speed with vehicle

speed can be plotted by inclusion of a plot statement. The result is shown in Figure

S3.20a. At speeds below 1.8 m/s the engine speeds are below 1000 rpm.

b) The engine rpm only affects the engine torque that is calculated in the MATLAB

function Fixed_thrt (see Figure 3.60). In other words, at the engine speed

corresponding to the vehicle speed, the engine torque is calculated and transferred to

the driving wheels. In order to limit the engine speed, including the following single

statement inside the function is sufficient:

if omega < 1000, omega=1000; end

However, when regenerating the engine speed in the main program from the values

of vehicle speed (output of the ode function), it is also necessary to include the

above statement inside the main program as well.

Inclusion of this limit on the engine speed has little effects on the results. Figure

S3.20b compares the results for the first 5 seconds for both cases with and without

the rpm limit.

Figure S3.20a Variation of engine speed with vehicle speed (No rpm limit)

Figure S3.20b Comparison of the output results with and without rpm limit

Problem 3.21

In a vehicle roll-out test on a level road the variation of forward speed with time is found to be of

the form:

, where a, b and d are three constants.

0 10 20 30 40 500

1000

2000

3000

4000

5000

6000

Vehicle speed (m/s)

Eng

ine

sp

ee

d (

rpm

)

0 1 2 3 4 50

5

10

15

20

Velo

city

(m/s

)

0 1 2 3 4 50

20

40

60

Time (s)

Dis

tan

ce (

m)

With rpm limit

Without rpm limit

)tan( dtbav

a) Assume an aerodynamic resistive force in the form of 2cvFA and derive an expression

for the rolling resistance force FRR.

b) Write an expression for the total resistive force acting on the vehicle.

Result: (b) FR = md (a+v2/a)

Solution:

a) From Equation 3.145 for the coast down:

2cvFdt

dvm RR

The rolling resistance force FRR, therefore, is:

2cvdt

dvmFRR

Differentiation from the given relation for the speed results in:

2va

dad

dt

dv

Substitution in equation for the rolling resistance force leads to:

2)(1

vacmda

madFRR

b) The total resistive force also includes the aerodynamic force. Thus:

222

1)(

1v

aamdcvvacmd

amadFFF ARRR

Problem 3.22

Two specific tests have been carried out on a vehicle with 1300 kg weight to determine the

resistive forces. In the first test on a level road and still air the vehicle reaches a maximum speed

of 195 km/h in gear 5. In the second test on a road with slope of 10%, the vehicle attains

maximum speed of 115 km/h in gear 4. In both tests the engine is working at WOT at 5000 rpm,

where the torque is 120 Nm.

a) If the efficiency of the driveline is 90% and 95% at gears 4 and 5 respectively, determine

the overall aerodynamic coefficient and the rolling resistance coefficient.

b) If the gearbox ratio at gear 5 is 0.711 and the wheel effective radius is 320 mm, assume a

slip of 2.5% at first test and determine the final drive ratio.

c) Calculate the ratio of gear 4 (ignore the wheel slip).

Results: (a) c=0.314, fR=0.014, (b) nf =4.24, (c) n4=1.206

Solution:

a) In both tests the vehicle attains steady state motion. Thus for gears 5 and gear 4

the equations of motion are:

5

2

555 )( vcvmgfP Red

44

2

4444 )sincos( vmgcvmgfP Red

As the engine working condition is identical for both cases, the engine powers Pe5

and Pe4 are equal to Tee. Thus the only two unknowns in the two above equations

are fR and c. The solution is:

cos

sincos

2

5

2

4

5

5

4

4

vv

mgvv

T

c

ddee

and mg

cvv

T

f

dee

R

2

5

5

5

For the given numerical values the results are obtained as c=0.3135 and fR=0.0143.

b) From Equation 3.137:

ww

xr

vS

1

where w = e/(nf n5). Substituting into the above equation results in the following

equation for nf:

xew

f Svn

rn 1

55

The numerical result is (use e in rad/s and Sx=0.025) nf=4.2418.

c) Since the wheel slip is ignored:

4

4vn

rn

f

ew

and the numerical result is n4= 1.2056.

Problem 3.23

For a vehicle with specifications given in table below, engine torque at WOT is of the following

form:

Te=100+a(e-1000)-b(e-1000)2 ,

a=0.04 , b= 810-5

, e < 6000 rpm

The driveline efficiency is approximated by 0.85+i/100 in which i is the gear number.

a) Determine the maximum engine power.

b) What is the maximum possible speed of the vehicle?

c) Calculate the maximum vehicle speed at gears 4 and 5.

Results: (a) 69,173 W, (b) 170.6 km/h, (c) 169.9 and 142.6 km/h

Table P3.23 Vehicle information

1 Vehicle mass 1200 kg

2 Rolling Resistance Coefficient 0.02

3 Tyre Rolling Radius 0.35 m

4 Final drive Ratio 3.5

5 Transmission Gear Ratio 1 4.00

6 Gear Ratio 2 2.63

7 Gear Ratio 3 1.73

8 Gear Ratio 4 1.14

9 Gear Ratio 5 0.75

10 Aerodynamic Coefficient CD 0.4

11 Frontal Area Af 2.0 m2

12 Air density A 1.2 kg/m3

Solution:

a) The maximum engine power is the maximum of the following function:

Pe=[100+a(e-1000)-b(e-1000)2] e

/30

Differentiation with respect to e and equating to zero results in:

01001010)40002(3 632 babab ee

which has a positive answer of e = 5,092.2 rpm. Substituting this value in the above

equation for the power gives: Pmax = 69,173 W.

b) The maximum possible speed of vehicle is achieved when the maximum power is

used to propel the vehicle. Therefore:

max

2

maxmax )( vcvmgfP Rd

in MATLAB the solution can be found from:

F0=m*g*fR; c=Cd*Af*air_dens/2;

vmax = 3.6*fsolve(@(x) Pmax*eta-(F0+c*x^2)*x, 20 ,optimset('Display','off'))

The answer is vmax=170.62 m/s.

c) According to the solution given in Example 3.7.3, Equation 3.112 for this problem

can be written in the form (driveline efficiency is also included):

0)( 3*

2

2*

1 kvkvk

with n

crbkk

d

w

21 , )2000(2 bakk ,

n

rFbak

d

w

063

3 1010100 and

wr

nk

30 .

The numerical results for 14.15.34 nn and 75.05.35 nn are 47.19 m/s

(169.9 km/h) and 39.62 m/s (142.6 km/h) respectively. Note that the maximum

speed in gear 4 is larger than that of gear 5. In addition the engine speeds are 5,137

and 2,837 rpm and are below 6000 rpm in both gears.

Problem 3.24

For the vehicle of Problem 3.23,

a) For a constant speed of 60 km/h over a slope of 10% which gears can be engaged?

b) For case (a) in which gear the input power is minimum?

c) On this slope what would be the maximum vehicle speed in each gear?

Hint: The following table is useful for solving this problem.

Table P3.24

Parameter Gear 1 Gear 2 Gear 3 Gear 3 Gear 5

1 Engine speed rpm

2 Engine torque Nm

3 Vehicle maximum speed km/h

Solution:

a) As the speed of motion is given (v*), the resistive forces are known quantities. In

order to have a constant speed, the condition is FT=FR=F*=known. For the specified

speed and traction force, each engaged gear will need a specific engine speed (

** vr

n

w

e ) and a specific engine torque ( we rn

FT

** ).

The maximum engine speed is bounded at 6000 rpm, therefore one condition for

each gear is to make engine turn at speeds below 6000 rpm. The other constraint is to

require engine torques below the maximum engine torque. Each gear that satisfies

these two conditions, can be engaged and produce the specified vehicle speed. The

first two rows of the proposed table are to check these two constraints. With the

following MATLAB commands the necessary information is obtained:

n =nf*[4.0 2.63 1.73 1.14 0.75]; % ratio of gears 1-5

j=1: 5;

d_eff=0.85+j/100; % Driveline efficieny array

wstar=30*n*vstar/rW/pi; % Engine speed in each gear

Te_star=rW*FR./(d_eff.*n); % Engine torque in each gear

The numerical results presented in the first two rows of Table S3.24, indicates that

only gears 2 and 3 satisfy the two constraints.

The following MATLAB commands can also be used to generate the results

automatically:

pos_w=find(wstar


c) The condition for having a dynamic balance is FT=FR, but the value of FR is not

known this time. This problem is exactly similar to the part (c) of Problem 3.23, the

only difference being the value of F0 which is:

)sincos(0 RfmgF

However, in this case a question arises of whether or not the high gears (e.g. gears 4

and 5) can be held at a constant speed on the slope? The answer to this question lies

with the solution of speed equation 0)( 3*

2

2*

1 kvkvk . In fact for those gears that

a dynamic balance point is not available, the roots of the equation will be complex

conjugates. In addition the result obtained for the speed must also comply with the

kinematic constraint of engine speed below 6000 rpm.

The following MATLAB program is based on the equations given in Problem 3.23

and above explanation. Running this program with proper data will automatically

specify the possible gears and their maximum speeds shown in the 3rd

row of the

Table S3.24. In some gears the maximum speed is its kinematic limit and in others

its dynamic balance point.

for i=1: length(n)

ni=n(i);

eta=d_eff(i);

k=ni*30/rW/pi;

k1=-(b*k^2+c*rW/ni/eta);

k2=k*(a+2000*b);

k3=100-1000*a-1e6*b-F0*rW/ni/eta;

vmax=roots([k1 k2 k3]);

ireal=isreal(vmax); % 0 if vmax is a complex number

vd(i)=ireal*vmax(1); % vd will be zero if vmax is complex

end

wd=30*n.*vd/rW/pi; % Engine speed in each gear

pos_wd=find(wd0); % Check for vehicle speeds to be > 0

gearnumber=intersect(pos_wd , pos_vd); % Acceptable gears that satisfy both conditions

for i=1: min(gearnumber)-1

vk(i)=6000*pi*rW/30/n(i); % Gear below 'x' reach their kinematic speed

end

i_acc=1: 1: gearnumber-1;

vi=[i_acc gearnumber; vk*3.6 vd(gearnumber)*3.6];

Problem 3.25

The vehicle of Problem 3.23 is moving on a level road at the presence of wind with velocity of

40 km/h. Assume CD=CD0+ 0.1 |sin |, in which is the wind direction relative to the vehicle

direction of travel. Determine the maximum vehicle speed in gear 4 for:

a) A headwind (=180)

b) A tailwind (=0)

c) A wind with =135 degree.

Results: (a) 142.5, (b) 192.0, (c) 140.5 km/h

Table S3.24

Parameter Gear 1 Gear 2 Gear 3 Gear 3 Gear 5

1 Engine speed 6366.2 4185.8 2753.4 1814.4 1193.7 rpm

2 Engine torque 44.74 67.3 101.1 151.7 228.0 Nm

3 Vehicle maximum speed 56.6 86.0 115.1 - - km/h

Solution:

This problem is essentially similar to the part (c) of Problem 3.23. The difference

here is the aerodynamic drag force which is different. It is defined as:

RA=0.5A (CD0+ 0.1 |sin |) AF 2

Av

The angle and the air speed vA are different for the three cases. The former is given

for each case but the latter must be determined. The air velocity vA is (see Figure

3.23 considering V=0 and W=):

jvivv WVWAsin)cos( v

The effect of wind direction on the aerodynamic force is already considered in CD,

and the air speed is the component of vA in the opposite direction of motion or,

cosWVA vvv

The equation to be solved in this example is similar to that of Problem 3.23 part (c),

but instead of c(v*)2 in the aerodynamic force, the term c(vA)

2 must be used. This will

only change k2 and k3:

cos2)2000(2 Wd

w vn

crbakk , 20633 )cos(1010100

W

d

w vcFn

rbak

The MATLAB program given in the solution of Problem 3.24, part (c) can be

modified for this problem. The basic changes are:

wd=[180 0 135]; % Wind direction array

Cd=Cd0+0.1*abs(sin(wd*pi/180)); % Cd array

C=Cd*Af*air_den/2;

vwd=vw*cos(wd*pi/180)/3.6; % Array of wind speed in the direction of motion

for i=1: length (wd) % Loop for each case

% Statements (modify the loop of Problem 3.24, part (c)) end

The results obtained from this program for the three cases are 142.5, 192.0 and 140.5

km/h.

Problem 3.26

Two similar vehicles with exactly equal properties are travelling on a level road but in opposite

directions. Their limit speeds are measured as v1 and v2 respectively. Engine torque at WOT is

approximated by following equation:

Te=150-1.1410-3

(-314.16)2

Determine the aerodynamic drag coefficient CD and wind speed in direction of travel vw by:

a) Writing a parametric tractive force equation in terms of vehicle speed for both vehicles.

b) Then write a parametric resistive force equation in terms of speed for both vehicles.

c) Equate the two equations for each vehicle and use the numerical values of Problem 3.23 for m,

fR, Af, A, rW and the additional information given in table below,

Results: 0.25 and 19.32 km/h

Table P3.26

1 Transmission Gear Ratio 0.9

2 v1 180

3 v2 200

Solution:

a) The tractive force for both vehicles is:

23- 314.16)-(101.14-150 W

e

W

Tr

nT

r

nF

In terms of the vehicle speed:

2i

3- 314.16)-(101.14-150 vr

n

r

nF

WW

T

in which vi is the speed of the ith vehicle.

b) The resistive forces for the first and second vehicles are,

2

11 )( wRR vvcmgfF

2

22 )( wRR vvcmgfF

c) At the limit speed, the tractive and resistive forces are equal and the governing

equations for the vehicles are:

2

1

2

1

3- )(314.16)-(101.14-150 wRWW

vvcmgfvr

n

r

n

2

2

2

2

3- )(314.16)-(101.14-150 wRWW

vvcmgfvr

n

r

n

With the given numerical values for v1 and v2, the above equations reduce to

1

2

1 )( kvvc w and 22

2 )( kvvc w . Defining 1

2

k

kr , the solutions for vw, c and

CD are: r

rvvvw

1

12 , 2

1

1

)( wvv

kc

and

fA

DA

cC

2

The numerical results are:

k1=925.24, k2=760.22, vw=5.37 (19.32 km/h), c=0.3018 and CD=0.252

Problem 3.27

While driving uphill in gear 4 on a road with constant slope , the vehicle of Problem 3.23

reaches its limit speed Uv at an engine speed of at a still air. The same vehicle is then driven

downhill on the same road in gear 5, while keeping the engine speed same as before. Engine

powers for uphill and downhill driving are UP and DP respectively. The tyre slip is roughly

estimated from equation Sx=S0+P10-2

(%) where S0 is a constant and P is power in hp.

Assume a small slope angle and use the additional data given in the table to determine:

a) Uphill and downhill driving speeds

b) Road slope

Results: (a) 116.9 and 165.1 km/h, (b) 9.4 %

Table P3.27

1 Uphill power PU 90 hp

2 Downhill power PD 10 hp

3 Tyre basic slip S0 2.0 %

4 Gear progression ratio n4/n5 C4 1.4

Solution:

a) In both cases the vehicle attains the limit speeds. Thus for gears 4 and 5 the

equations of motion are:

UURUd vmgcvmgfP )sincos(2

4

DDRDd vmgcvmgfP )sincos(2

5

In contrast with the case of Problem 3.22 the engine working condition is not

identical for both cases. In fact in this problem in spite of having equal engine

speeds, the throttle inputs are not equal and the engine powers Pu and Pd are different

as given. Thus the three unknowns in the two above equations are vU, vD and .

Assuming a small slope angle (i.e. cos =1 and sin = in rad) simplifies the

equations and can be solved to obtain:

UDRUDdDUdUDDU vmgvfvPvPvvvvc 2)( 5422

An additional equation can be obtained from the information given for the tyre slips.

The vehicle speed is related to the slip by:

)1( ii

wi S

n

rv

The ratio of speeds in gears 5 and 4 are ( is identical for both):

4

54

45

54

1

1

1

1

S

SC

Sn

Snk

v

v

U

D

Substituting into the equation obtained earlier leads to an equation for vD:

02)1

1( 543

2 DdUdDRD PkPmgvfv

kc

vU and can then be determined from:

k

vv DU and RU

U

Ud fcvv

P

mg

24

1 (rad)

Problem 3.28

For the vehicle of Problem 3.23,

a) Derive a general parametric expression for the value of speed v* at the maximum

attainable acceleration.

b) Use the numerical values and determine the values of v* at each gear.

c) Calculate the maximum accelerations at each gear.

Results: (a)

)30

(2

)2000(30

3

3

2

22

2*

W

id

d

W

i

r

nbc

ba

r

nv

, (b) 9.06, 13.38, 18.51, 21.45 and 17.88 m/s, (c) 4.07,

2.59, 1.55, 0.796 and 0.298 m/s2.


Numerical values are:

k = 1.4115, vD= 45.85 (165.1 km/h), vU= 32.48 (116.9 km/h) and = 0.093 (rad) or

5.34 (deg) or 9.4%.

Solution:

a) The vehicle acceleration in each gear ni is:

)(1 2cvmgfT

r

n

mm

FFa Red

W

iRT

Assuming no slip, Te in the above equation can be written as a function of the vehicle

speed:

2

100030

100030

100

v

r

nbv

r

naT

W

i

W

ie

Substituting into the above equation makes the acceleration a function merely of the

vehicle speed. Differentiation with respect to v and equating to zero results in:

02100030

230

cvv

r

nba

r

n

r

n

W

i

W

id

W

The solution to this equation is v* that maximises the acceleration:

2

2

*

3022

)2000(30

W

id

W

i

W

id

r

n

r

bnc

bar

n

v

b) The numerical results for v* are 9.06, 13.38, 18.51, 21.45 and 17.88 (m/s) for

gears 1-5. To obtain values of amax, the values for Te can be determined first and then

substituted in the first equation. The results are:

Te: 149.99, 149.84, 148.44, 139.14 and 110.60 Nm.

a: 4.07, 2.59, 1.55, 0.796 and 0.298 m/s2.

It should be noted that the answer for gear 5 is not valid since the engine speed drops

to 281 rpm.

Problem 3.29

In Section 3.9 the effect of rotating masses were discussed and equations for including this effect

in the acceleration performance of a vehicle were developed. From an energy consumption point

of view, when vehicle is accelerated to the speed of v, the rotating inertias will be at rotational

speeds related to v (ignore the tyre slip).

a) Write the kinetic energies for the vehicle body mass m and rotating masses Ie, Ig and Iw

b) From the kinematic relations, relate the rotational speeds to the vehicle speed

c) Write the energy terms in terms of vehicle speed v

d) Write the total energy of vehicle as: 25.0 vmE eqt

e) Determine the equivalent mass meq and compare it with Equation 3.130

Solution:

a) The kinetic energies for the vehicle body and rotating masses are:

25.0 mvEv , 25.0 eee IE ,

25.0 ggg IE and 25.0 www IE

b) From the kinematic relations, the rotational speeds can be written as:

g

eg

n

,

f

g

wn

The vehicle speed can be related to wheel rotational speed if no slip is assumed:

wwrv

c) Substituting the kinematic relations into the energy equations, all energy terms can

be expressed in terms of the vehicle speed v:

22)(5.0 vr

nIE

w

ee , 22)(5.0 v

r

nIE

w

f

gg and 22)

1(5.0 vr

IEw

ww

d) The total energy simply is the summation of all energy terms:

25.0 vmEEEEE eqwgevt

e) Substituting the energy terms into the above equation results in:

22222222 )1

(5.0)(5.0)(5.05.05.0 vr

Ivr

nIv

r

nImvvm

w

w

w

f

g

w

eeq

which simplifies to:

)](1

1[ 222 egfAw

w

eq InInIImr

mm

that is, exactly similar to Equation 3.130.

Problem 3.30

For a tyre with the Magic Formula information given in Table 3.3,

a) Plot the longitudinal force (F) against slip (s) for both traction and brake regions at

normal load values 1.0, 2.0, 3.0 and 4.0 kN (all in a single figure)

b) Plot coefficients of tyre-road friction for case (a)

c) At slip ratios 5, 10, 20 and 50%, plot the variation of Fx versus Fz (max Fz=5kN).

d) Differentiate the Magic Formula with respect to slip to find the value of slip at which the

force is maximum. Verify your results by comparing them with those of case (a).

e) In order to have an impression of the influence of different factors in the Magic Formula

tyre model, try the following for the above tyre in (a) at a normal load of 3.0 kN:

I. Multiply coefficient B by 0.8, 1.0 and 1.2 while keeping the other coefficients

unchanged. Plot all three results in a single figure.

II. Repeat I for coefficient C.

III. Repeat I for coefficient D.

IV. Repeat I for coefficient E.

Solution:

a) This part is similar to Example 3.3.1 but for the full range of slip variations. The

MATLAB program of Figure 3.12 can be modified for this part. The result is plotted

in Figure S3.30a.

b) This part is similar to Example 3.3.2. By dividing the longitudinal forces by the

normal load the coefficient of adhesion is obtained. The result for the whole range is

shown in Figure S3.30b.

c) The following MATLAB commands may be used to generate the results shown in

Figure S3.30c:

figure, hold on

sx=[5 10 20 50];

fz=0: 25: 5000;

for i=1: length(sx)

for j=1: 201

fx(j)=FX(fz(j), sx(i));

end

plot(fz, fx)

end

Note the nonlinear dependency of the longitudinal force on the slip.

d) The differentiation of Fx with respect to Sx results in the following equation:

)]([cos)(1

1)(1 22

BArctanC

BS

EE

B

BCD

S

F

xx

x

The above derivative can have two possible answers when equated to zero. These

correspond to the last two factors. Equating the second factor to zero results in:

1

1)( 2

EBS x

Since E is always smaller than unity, the above equation has no practical solution.

Thus the only possible answer results from:

0)]([cos BArctanC

-100 -50 0 50 100-5000

-4000

-3000

-2000

-1000

0

1000

2000

3000

4000

5000

Longitudinal slip (%)

Lon

gitud

ina

l fo

rce

(N

)

Fz=1.0 kN

Fz=2.0 kN

Fz=3.0 kN

Fz=4.0 kN


The solution of the above equation must satisfy:

CSBArctanEE1SBB xx

2tan)()(

This is a nonlinear equation for Sx that can be solved by the MATLAB command:

sx_star=fsolve(@(x) b*(1-e)*x+e*atan(b*x)-tan(pi/2/c), 1, optimset('Display','off'))

For the values of Fz=1000, 2000, 3000 and 4000 N, the answers are Sx= 12.95, 11.86,

11.03 and 10.35%. These values can also be found from the solution of the part (a).

From Figure S3.30a the approximate values are observed.

e) For this part, the function fx of Figure 3.12 must be modified too. This can be

done in different ways. The multipliers can be multiplied inside the function for each

of the coefficients at a time, or a general multiplier can be used that could be

controlled within the main program. The results for the parts I through IV are shown

in Figures S3.30e to S3.30g.

Figure S3.30a Variation of tyre longitudinal force with slip at different normal loads

Figure S3.30b Variation of adhesion coefficient with slip at different normal loads

Figure S3.30c Variation of tyre longitudinal force with normal load at different slips

-100 -50 0 50 100-1.2

-0.8

-0.4

0

0.4

0.8

1.2


Adh

esio

n c

oeffic

ient

1.0 kN

2.0 kN

3.0 kN

4.0 kN

0 1000 2000 3000 4000 50000

1000

2000

3000

4000

5000

6000

Normal load (N)

Lon

gitud

ina

l fo

rce

(N

)

Sx=5%

Sx=10%

Sx=20%

Sx=50%

Figure S3.30d Effect of the Magic Formulas B factor on the Fx

Figure S3.30e Effect of the Magic Formulas C factor on the Fx

-100 -50 0 50 100-4000

-3000

-2000

-1000

0

1000

2000

3000

4000


Lon

gitud

ina

l fo

rce

(N

)

Multiplier=0.8

Multiplier=1.0

Multiplier=1.2

-100 -50 0 50 100-4000

-3000

-2000

-1000

0

1000

2000

3000

4000


Lon

gitud

ina

l fo

rce

(N

)

Multiplier=0.8

Multiplier=1.0

Multiplier=1.2

Figure S3.30f Effect of the Magic Formulas D factor on the Fx

Figure S3.30g Effect of the Magic Formulas E factor on the Fx

-100 -50 0 50 100-4000

-3000

-2000

-1000

0

1000

2000

3000

4000


Lon

gitud

ina

l fo

rce

(N

)

Multiplier=0.8

Multiplier=1.0

Multiplier=1.2

-100 -50 0 50 100-4000

-3000

-2000

-1000

0

1000

2000

3000

4000


Lon

gitud

ina

l fo

rce

(N

)

Multiplier=0.8

Multiplier=1.0

Multiplier=1.2

Problem 3.31

The vehicle of Problem 3.23 is moving with a constant speed of 100 km/h. Use the tyre data of

Table 3.3 for each of the two driving wheels and for a front/rear weight distribution of 60/40,

determine

a) Longitudinal slip (in percentage) of the tyres for both cases of FWD and RWD

b) Repeat (a) for a 5-degree grade (ignore the load transfer)

c) Repeat (a) for a level road with adhesion coefficient of 0.4

Solution:

a) The vehicle is moving at a constant speed; nonetheless, the driving wheels are

exerting tractive force to overcome the resistive forces. The total tractive force

generated by the two driving wheels is equal to the total resistive forces. On a level

road the total resistive force is:

NcvmgfF RR 9.60578.2724.02.15.002.081.9120022

Each of the driving wheels produces half of the force (i.e. 302.95 N). Due to the

constant speed there is no longitudinal load transfer and the tyre loads are static

loads. For a FWD vehicle the loads on the driving wheels are 60% of vehicle weight

whereas for a RWD vehicle the load is 40%. Therefore for the FWD case, the load

on each tyre is 35323.081.91200 N and for the RWD case it is 2354 N.

Once the longitudinal and normal loads of a tyre are known, its slip can be

determined from the Magic Formula. However, the process is of a trial and error

nature and MATLABs fsolve is useful. The following program uses the Magic

Formula equations and determines the slip if the normal load is given for each case:

fz=??

c=1.65d0;

d=-21.3d-6*fz*fz+1144.d-3*fz;

e=-.006d-6*fz*fz+.056d-3*fz+.486d0;

bcd=(49.6d-6*fz*fz+226.d-3*fz)*exp(-.069d-3*fz);

b=bcd/(c*d);

sx=fsolve(@(x) 302.95-d*sin(c*atan(b*(1.d0-e)*x+e*atan(b*x)/b)), 2, optimset('Display','off'))

The answers for the FWD and RWD are 0.27 and 0.44% respectively.

b) In this case the resistive load increases and the normal loads also decrease (the

load transfer due to the slope is ignored), otherwise the solution method is similar.

The total resistive force is:

NcvfmgF RR 1631)sincos(2

Normal loads on front and rear tyres are cos 5 (0.996) times the previous loads.


Again, the above program with the new numerical values, generates slips of 0.75 and

1.24% for the FWD and RWD.

c) On a slippery road the tyre slip must be larger in order to produce a similar force

to that on a dry road. In order to change the adhesion coefficient of the Magic

Formula, the simplest way is to multiply the coefficient D by a factor of less than

unity. Assuming that this factor is the adhesion coefficient and is 1 for the dry

road, for a road with the adhesion coefficient of 0.4, the multiplier should be 0.4.

The solution will then be exactly similar to that of part (a) with exception that a

factor of 0.4 is considered for d in fsolve function. The answers for FWD and

RWD cases are 0.69% and 1.14%.

Problem 4.1

Explain why the term Nf+Nr of Equations 4.19 and 4.20 is not necessarily equal to Wcos and

discuss the conditions of equality.

Solution:

According to Equations 4.19 and 4.20, Nf and Nr each belong to FWD and RWD

categories. Summing up the two values, therefore, does not necessarily have a

physical meaning.

However, in order to find out the conditions of equality, using Equations 4.19 and

4.20 one obtains:

cos)( WkkNN RFrf

Thus for Nf+Nr= Wcos from Equation 4.25:

1

hl

hfa

hl

hfb

p

R

p

R

(*)

The solution for p is:

l

h

l

a

l

bfRp /)(2 (**)

Apart from the unimportant parameter fR, only 3 dimensionless parameters p , a/l

and h/l are involved. For the practical values, the plots of RF kk are shown in

Figures S4.1a and S4.1b for the variation of p. Figure S4.1a is drawn at h/l =0.3 for

different values of a/l and it is observed that onl

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[David Crolla_Behrooz Mashadi]Vehicle Powertrain Systems-Integration and Optimization_Solution Manual