problem 3...b) for a vehicle of 1000 kg mass and total rolling resistance force of 200 n, when...

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

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%?

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

22

0 , P

vvmS

3

)( 3

0

3 , (b) 46.3, 57.9 and 77.2 kW, (c) 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?

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

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)

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.

0 2 4 6 8 100

5

10

15

20

25

30

Time (s)

Velo

city

(m/s

)

LS model

NRF model

Figure S3.7 Power requirements for different acceleration times

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.

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

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([ 11

2

10

2

10 vpvvFPpmFCS , (c) )5.0ln( 11

2

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

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.

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.

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

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 ivvi

and

)1(max0 iSSi

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.

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)

0 5 10 15 20 250

5

10

15

20

25

30

35

Time (s)

Spe

ed

(m

/s)

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

Problem 3.14

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

0 5 10 15 20 250

100

200

300

400

500

600

Time (s)

Dis

tan

ce (

m)

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

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<5 % No delay after gear 5!

t0=max(t);

tf=t0+tdelay;

x0=[v(end) s(end)];

p=[0 0 0]; % No traction force

[t,x]=ode45(@Fixed_thrt, [t0 tf], x0);

v=x(:,1);

s=x(:,2);

end

% Now plot the results

p=[p1 p2 p3 p4]; % Set back the engine torque

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

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)


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

)

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.



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

Time (s)

Acc

ele

ration

(m

/s2

)

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?

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

0 10 20 30 40 500

1000

2000

3000

4000

5000

6000

Vehicle speed (m/s)

Eng

ine

sp

ee

d (

rpm

)

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.

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)

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

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

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= 8×10-5

, e < 6000 rpm

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

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

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

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.

11 Frontal Area Af 2.0 m2

12 Air density A 1.2 kg/m3

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

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

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=150-1.14×10-3

(-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,

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

Results: 0.25 and 19.32 km/h

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 %

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.

Table P3.26

1 Transmission Gear Ratio 0.9

2 v1 180

3 v2 200

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

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.

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

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.

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

-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

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 Formula’s B factor on the Fx

Figure S3.30e Effect of the Magic Formula’s 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 Formula’s D factor on the Fx

Figure S3.30g Effect of the Magic Formula’s 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

problem 3...b) for a vehicle of 1000 kg mass and total rolling resistance force of 200 n, when...

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