Car Modeling and Simulation

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Table of contents Introduction………………………………………………….….2 Model of the system…………………………………………..3 Model of the bump……………………………………………..4 Differential Equation of Motion (2DOF)…………………….6 Natural Frequency and Mode Shapes……………………...8 General Solution……………………………………………...19 D.E of Motion using Lagrange Equation…………………20 Differential Equation of Motion (4DOF)…………………..22 Nonlinear Differential Equation of Motion……………….24 Solution with Laplace Method……………………………..25 Numerical Solution…………………………………………..26 Sample of Excel worksheet………………………………...31 VBA code for numerical solutions………………………..32

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Introduction

In this project, I’m trying to model a car which could be a Sedan

car, a truck, or an SUV, when passing through a Bump. This bump

is acting like a base excitation for our system.

First of all we will assume a 2DOF linear system; however, I will

derive the differential equation for 4DOF and Non linear system.

The different parts of this project are:

1. Modeling of the system

2. Modeling of the bump

3. Differential Equation of Motion 2DOF (linear)

4. Calculation of Natural Frequency and Mode shapes

5. Differential Equation of Motion Using Lagrange Eq.

6. Differential Equation of Motion 4DOF (linear)

7. Differential Equation of Motion 2DOF(Non linear)

8. Calculating X(t),θ(t) with Laplace Method

9. Calculating X(t),θ(t) with Numerical method Runge Kutta(4)

10. State variables for 2DOF (linear)

11. State variables for 2DOF (non linear)

12. VBA codes of 4-th order Runge Kutta for this problem

13. Graphs of X(t),θ(t) with some inputs (with Excel)

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Modeling of the System

Model of a car:

• First assumption: The effects of the tires are negligible.

• Second assumption: The left side and the right side of the car are symmetric.

Dynamic model:

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Modeling of the Bump

• First assumption: We assume that the bump is like a half sine wave.

So; the effect of bump is like base excitation on the system: z1, z2.

• Second assumption: We assume that the car’s speed is constant (acceleration is

zero) at this period.

SpeedCarVV

Dt

D

dhz p

tVd :

.sin.

1

.

01 = →��

��

�= =π

��

�

��

��

�

=

,0

.sin.0

1 D

dh

z

π

1

1

tptif

tptif

≥→

<→

��

��

��

��

�=

0

..sin.

0

02D

tVhz

π

2

2

tptif

tpttif

ttif

l

l

>→

<<→

<→

Note: V

DLtp

V

Ltl

+=→= 2

Top graph: Front wheel

h: height of the bump

d: horizontal coordinate

D: length of the bump

L: distance between the front and back tires

Bottom graph: rear wheel

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Other form of z1, z2 function:

Introduce �

=1

0)(tU c

Ct

Ct

>

≤

( )

[ ]��

��

��

��

�−=

��

��

�−=

�

D

tVhtUtUz

D

tVhtUz

tptl

tp

..sin..)()(

..sin..)(1

0

02

0

01

2

1

π

π

Check,

0..

sin.).11(

..sin.

..sin.).01(

0..

sin.).00(

..sin.).01(

0

022

0

0

0

022

002

0

011

=��

��

�−=→>

��

��

�=�

�

��

�−→≤≤

=��

��

�−=→<

��

��

�−=→<

D

tVhztpt

D

tVh

D

tVhztptt

D

tVhztt

D

tVhztpt

L

L

π

ππ

π

π

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Differential Equation of Motion 2DOF (linear)

Assume: θ

θ

21

22

lxz

lxz

+<

−<

� =..

. xmFx

..

1122 xmFsFcFcFs =−−−−

)(

)(

2111

2122

zlxkFs

zlxkFs

−−=

−−=

θ

θ

)(

)(

2

..

1

.

11

2

..

2

.

22

zlxcFc

zlxcFc

−−=

−−=

θ

θ

..

1

..

1

..

11

.

1111112

.

2

.

22

.

222222 xmzclcxczklkxkzclcxczklkxk =+−−+−−++−++−� θθθθ

22112

.

21

.

1221121

.

2211

.

21

..

)()()()( zkzkzczclklkxkklclcxccxm +++=−+++−+++� θθ

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� =..

θIMcG

..

11112222 .. θIlFslFclFclFs =−−+

..

11111

..

1

.

112

..

2

.

222222 )()()()( θθθθθ Izlxlkzlxlczlxlczlxlk =−+−−+−−−+−−

zlkzlkzlczlcxlklklklkxlclclclcI 221112

.

221

.

112211

2

11

2

22

.

2211

.2

11

2

22

..

)()()()( −+−=−+++−+++� θθθ

If we put 1, 2 in matrix format;

�

��

� �

��

�

+−

−++

�

�

��

�

� �

��

�

+−

−+

�

�

��

�

� �

��

�

θθθ

x

lklklklk

lklkkkx

lclclclc

lclccx

I

m...

0

02

11

2

222211

221121.

.

2

11

2

222211

22111..

..

�

�

��

�

�

−+−

+++=

2221112

.

221

.

11

22112

.

21

.

1

zlkzlkzlczlc

zkzkzczc

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Calculation of Natural Frequency and Mode shapes

Assuming Reasonable values for all parameter, we have:

2.rmI = r : Radius gyration of the car

Reasonable approximate values for a Sedan car;

22 6.0 mr = kgm 1000= msNcc .50021 ==

mNkk 500021 == ml 8.01 = ml 2.12 =

Reasonable approximate values for a SUV;

22 65.0 mr = kgm 3000= msNcc .150021 ==

mNkk 150021 == ml 11 = ml 5.12 =

Solved for a bump with mh 1.0= , mD 5.1=

Eigen Value Problem and Mode Shapes

For our model we assume c1 = c2 = c and k1 = k2 = k

�

��

�=

�

��

� �

��

�

+−

−+

�

�

��

�

� �

��

�

+−

−+

�

�

��

�

� �

��

�→

)(

)(.

)()(

)(2.

)()(

)(2.

0

0

2

1

2

1

2

121

21.

.

2

1

2

121

21..

..

3.

tg

tgx

llkllk

llkkx

llcllc

llccx

I

meq

θθθ

Where:

[ ] [ ][ ] [ ] )cos.,sin..()()(.)cos.,sin..()(1.)(

)cos.,sin..()()()cos.,sin..()(1)(

21

21

20102

001

tctktutulhtctktulhtg

tctktutuhtctktuhtg

ffftptLffftp

ffftptLffftp

ωωωωωω

ωωωωωω

+−++−=

+−++−=

Note: D

Vf

.πω =

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Eigen Value Problem:

[ ] [ ]( ) 0.2 =− mk ω

0.)()(

)(2det

22

2

2

121

21

2

= �

��

�

+−

−−�

Illkllk

llkk

ω

ω

0.

)(.)(

22

21

222

2

2

1

4 =+

+ �

��

�++−�

IM

llkll

I

k

m

kωω

Im

llkll

I

k

m

kll

I

k

m

k

.

)(4)(

2)(

22

21

22

2

2

2

1

2

2

2

1

2 +− �

��

�++±++=�ω

For a sedan car we have:

kgm 1000= , 2.rmI = and 22 6006.0 NmIr =�=

8.01 =l 2.12 =l , mNkk 500021 == msNcc .50021 ==

If we plug in eq.(5)

06001000

)2.18.0.()5000()2.18.0(

600

5000

1000

10000 222224 =

⋅

++ �

��

�++− ωϖ

0677.166334.27 24 =+−� ϖω \

182.92

1 =�ω 151.182

2 =ω

)sec(03.31 rad=�ω )sec(26.42 rad=ω

Back sub �1 in (4)

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).(409.0

1

8.48902000

2000818

)600).(182.9()2.18.0.(5000)2.18.0.(5000

)2.18.0.(5000)1000).(182.9(10000

.)()(

)(2

1

222

1

2

2

2

121

21

2

1

ModeFirst

Illkllk

llkmk

→ �

��

�=�

�

��

�

−

−=

�

��

�

−+−

−−=

�

��

�

−+−

−−�

ϕ

ω

ω

Sub �2 in (4)

).(075.4

1

6.4902000

20008151

)600).(151.18()2.18.0.(5000)2.18.0.(5000

)2.18.0.(5000)1000).(151.18(10000

2

22

ModeSecond→ �

��

�

−�

�

��

�

−−

−−=

�

��

�

−+−

−−

ϕ

∴ Approximate Mode Shapes and Natural Frequencies

For Sedan Cars are,

�

��

�

−=�=

�

��

�=�=

075.4

1sec26.4

409.0

1sec03.3

22

11

ϕω

ϕω

rad

rad

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For an SUV we have

kgm 3000= 2/ rmI = and 2/192064.0 mNIr =→=

mNkk 1500021 == 11 =l 5.12 =l

mNcc 150021 ==

Plug the numbers in (5),

0)1920).(3000(

)5.11.()15000(.5.11(

1920

15000

3000

30000 222224 =

++ �

��

�++− ωω

014.24439.35 24 =+−� ωω

39.92

1 =→ ω , 262

2 =ω

sec064.31 rad=�ω , sec1.52 rad=ω

Back sub ω 1 in (4),

�

��

�

−

−=

�

��

�

−+−

−−

2.307217500

75001830

)1920).(39.9()5.11.(15000)5.11.(15000

)5.11.(15000)3000).(39.9(300002

�

��

�=→

244.0

11ϕ

Back sub �2 in (4),

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�

��

�

−−

−−=

�

��

�

−+−

−−

11707500

750048000

)1920).(26()5.11.(15000)5.11.(15000

)5.11.(15000)3000).(26(300002

�

��

�

−=→

4.6

12ϕ

∴ Approximate Mode Shapes and Natural Frequencies

For an SUV is:

�

��

�

−=�=

�

��

�=�=

4.6

1sec1.5

244.0

1sec06.3

22

11

ϕω

ϕω

rad

rad

Conclude:

If we vibrate a sedan and a SUV at their second Natural Frequency, the SUV is more

capable of rotating than a sedan. However, at their first Natural Frequency, the situation

is vice versa.

Important Note:

[ ] �

��

�

+−

−=

�

��

�

+−

−=

2

1

2

121

21

2

1

2

121

21 2

).().(

).(.2

llll

llk

llkllk

llkkk

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[ ] �

��

�

+−

−=

�

��

�

+−

−=

2

1

2

121

21

2

1

2

121

21 2.

).().(

).(.2

llll

llC

llCllC

llCCC

As we can see above, the [C] matrix is a factor of [k]; so we will be able to calculate

modal C.

Modal Mass: (for sedan car)

�

��

�

−=

075.4409.0

11ϕ ,

�

��

�=

6000

01000M

[ ] [ ][ ] �

��

�==

�

��

�

109630

01100..

0

0

2

1 ϕϕ mm

m T

kgm 11001 =� , kgm 109632 =

Modal Damping: (sedan)

[ ] [ ][ ] [ ] �

��

�

−

−==

�

��

�

1090200

2001000,..

0

0

2

1CC

C

C Tϕϕ

�

��

�=

�

��

��

199000

01010

0

0

2

1

C

C

msNC .10101 =� , msNC .199002 =

Modal Stiffness: (sedan)

[ ] [ ][ ] [ ] �

��

�

−

−==

�

��

�

104002000

200010000,..

0

0

2

1km

k

k Tϕϕ

�

��

�=

�

��

��

1990000

010100

0

0

2

1

k

k

mNk .101001 = mNk .1990002 =

∴ Our uncoupled System is,

[ ] �

��

�=

�

��

�

2

1.

η

ηϕ

θ

x

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��

�

=++

=++

)_(mod..

)_(mod..

2222

.

22

..

2

111

.

11

..

11

forcealFkCm

forcealFkCm

ηηη

ηηη

m1, c1, k1, m2, k2, c2, are already calculated. We should find only F1 and F2 which are

model forces.

Modal Damping: (For SUV Car)

[ ] �

��

�

−

−=

�

��

�

+−

−=

4875750

7503000

).().(

).(.2'

2

1

2

121

21

llCllC

llCCC

[ ] [ ][ ] �

��

�==

�

��

�

212800

02920'.'.'

0

0'

2

'

1 ϕϕ CC

C T

msNC .2920'

1 = , msNC .212280'

2 =

Modal Stiffness: (For SUV Car)

[ ] �

��

�

−

−=

�

��

�

+−

−=

487507500

750030000

).().(

).(.2'

2

1

2

121

21

llkllk

llkkk

�

��

�=

�

��

�

2128000

029200

0

0'

2

'

1

k

k

mNk .29200'

1 =� , mNk .2122800'

2 =

∴ Our uncoupled system is:

[ ] �

��

�=

�

��

�

2

1.

η

ηϕ

ϕ

x

��

�

=++

=++

)_(mod..

)_(mod..

'

2

'

2

.'

2

..'

2

'

1

'

1

.'

1

..'

1

forcealFkCm

forcealFkCm

ηηη

ηηη

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Calculation of Modal forces:

[ ] (6) (t)g

(t)g�

F

F

2

1T

2

1

�

��

�= �

��

�

Finding g1 and g2

f

tPtLtPtL

tPtA

D

v

D

vttUtU

D

vz

D

vttUtUz

D

vttU

D

vz

D

vttUz

zkzkzczctg

tgtg

ωπ

πππ

πππ

=

��

��

�−=→�

�

��

�−=

��

��

�−=→�

�

��

�−=

+++=

cosh )]()([ sinh )]()([

cosh )](1[ sinh )](1[

)(

)( and )( Calculate

o2o2

o1o1

221122111

21

22

1

�

�

��

V

DL

V

L

V

D

xBy

twwctwtUtUltwwctwktUltg

zlkzlczlkzlctgThe

twwctwtUtUtwwctwktUtg

twtUtUtwtUktwtUtUwctwtUwctg

ffftPtLffftP

ffftPtLffftP

ftPtLftPiftPtLfftPf

+===

+−−+−=

−−+=

+−++−=

−+−+−+−=

21

21

21

2121

PLP

222o111o2

2222221111112

22o11o1

o2oo2o11

t, t, t: codesin that Note

, , for x, solution numerical a have will weMethod, Kutta Rungein CodesVBA using

)cossink)](()([h)cossin)]((1[h)(

)(for method same

)cossink)](()([h)cossin)]((1[h)(

sinsin)]()([hksin)](1[hcos)]()([hcos )](1[h)(

θθ ��

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500c 5000,K , m 22.18.0L L and

bumb) theoflength and(height 1.5D m 1.0h with ) (6in put them and )( and )( calculate should weSo

21

o21

===+=+=

==

L

tgtg

sec 0.52 t sec 522.0t

sec 3.0 t sec 298.07.6

2t

sec 22.0t sec 224.07.6

5.1t

rad/sec 03.145.1

7.6 m/sec 6.7mph 15V 1)

22

11

PP

LL

PP

f

=≈=+

=

=≈===

=≈===

=×

==→==

v

DL

L

v

v

D

D

v ππω

z1, z2 vs. time

-0.02

0

0.02

0.04

0.06

0.08

0.1

0.12

0 0.1 0.2 0.3 0.4 0.5 0.6

time

z1, z2 (heig

ht, m

ete

rs)

z1

z2

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sec 0.29 t sec 292.012

5.3t

sec 17.0 t sec 167.012

2t

sec 12.0t sec 125.012

5.1t

rad/sec 13.255.1

12 w m/sec 12 mph 27 v)2

)]03.14cos(8.841)03.14sin(600)][()([)]03.14cos(2.562)03.14sin(400)][(1[)(

)]03.14cos(5.701)03.14sin(500)][()([)]03.14cos(5.701)03.14sin(500)][(1[ )(

formula, into numbers eInsert th

22

11

PP

LL

PP

f

52.03.022.02

52.03.022.01

=≈==+

=

=≈===

=≈===

=×

==→==

+−++−=

+−++−=

v

DL

L

v

v

D

D

v

tttUtUtttUtg

tttUtUtttUtg

ππ

With this velocity the graphs for )(&)( 21 tgtg are:

g1, g2 vs. time

-2000

-1500

-1000

-500

0

500

1000

1500

0 0.1 0.2 0.3 0.4 0.5 0.6

time

g1, g2

g1

g2

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sec 0.29 t sec 292.012

5.3t

sec 17.0 t sec 167.012

2t

sec 12.0t sec 125.012

5.1t

rad/sec 13.255.1

12 w m/sec 12 mph 27 v)2

22

11

PP

LL

PP

f

=≈==+

=

=≈===

=≈===

=×

==→==

v

DL

L

v

v

D

D

v ππ

)]44cos(2640)44sin(600)][()([)]44cos(1760)44sin(400)][(1[)(

)]44cos(2200)44sin(500)][()([)]44cos(2200)44sin(500)][(1[ )(

sec 0.16 tsec, 09.0 tsec, 07.0t

rad/sec44 wm/sec 21 mph 47 v)3

)]15.25cos(8.1507)13.25sin(600)][()([)]13.25cos(2.1005)13.25sin(400)][(1[)(

)]13.25cos(5.1256)13.25sin(500)][()([)]13.25cos(5.1256)13.25sin(500)][(1[ )(

formula, intonumber heInsert t

16.009.007.02

16.009.007.01

PLP

f

29.017.012.02

29.017.012.01

21

tttUtUtttUtg

tttUtUtttUtg

D

v

tttUtUtttUtg

tttUtUtttUtg

+−++−=

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Equation" Uncoupled" have we, cases allFor

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sin)()()coscos()(

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zero initial condition Or Force (displacement) excitation

By taking an inverse, we will find the solution; however this method usually

is not easy.

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Runge Kutta (4)

The general format of Runge Kutta method is based on state variables;

however the formulation of this method is provided:

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22 )(,4)(,3)(,2)(,1

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2var:

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32,3

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=

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1var:

1var:

1var:

1var:

41,4

31,3

21,2

11,1

iablestatekk

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y y , y

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y y y

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)( :

))()()(

)()()()(

==

=++−+++−+

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θθ

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In this case we have four state variables

AMatrix

)(

0

)(

0

y

y

y

- )(

- )(

- )(

-

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)(

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(Note: the related code is available in appendix)

linear silution v=6.7 m/sec runge kutta method

-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0 0.1 0.2 0.3 0.4 0.5 0.6

time

x

Series1

linear solution with runge kutta v=6.7 m/sec

-0.15

-0.1

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θθθθθθθθθ

θθθθ

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sin)()()coscos()( )1

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2

22

2

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11

22

22

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y 3) y

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VBA codes:

Option Base 1 Function u_tp1(t_in, t_p1)

If t_in >= t_p1 Then

u_tp1 = 1

ElseIf t_in < t_p1 Then

u_tp1 = 0

End If

End Function

Function u_tp2(t_in, t_p2)

If t_in >= t_p2 Then

u_tp2 = 1

ElseIf t_in < t_p2 Then

u_tp2 = 0

End If

End Function

Function u_tL(t_in, t_L)

If t_in > t_L Then

u_tL = 1

ElseIf t_in <= t_L Then

u_tL = 0

End If

End Function

Function u_tp(t_in, t_p1)

If t_in > t_p1 Then

u_tp = 1

ElseIf t_in <= t_p1 Then

u_tp = 0

End If

End Function

Function g_1(array_in, xin, t, dt)

Dim c_1 As Double, c_2 As Double ' damping coefficients

Dim k_1 As Double, k_2 As Double ' spring constants for suspension

k_1 = array_in(1, 1) ' spring constants for suspension

k_2 = array_in(2, 1)

c_1 = array_in(3, 1) ' damping coefficients

c_2 = array_in(4, 1)

g_1 = c_1 * z_dot1(array_in, xin, t, dt) + c_2 * z_dot2(array_in, xin, t, dt) + _

k_1 * z_1(array_in, xin, t, dt) + k_2 * z_2(array_in, xin, t, dt)

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End Function

Function g_2(array_in, xin, t, dt)



Dim l_1 As Double, l_2 As Double





l_1 = array_in(11, 1)

l_2 = array_in(12, 1)

g_2 = c_1 * l_1 * z_dot1(array_in, xin, t, dt) - (c_2 * l_2 * z_dot2(array_in, xin, t, dt)) + _

k_1 * l_1 * z_1(array_in, xin, t, dt) - (l_2 * k_2 * z_2(array_in, xin, t, dt))

End Function

Function z_1(array_source, xin, t, dt)

Dim velocity As Double

Dim time As Double

Dim wheelbase As Double, bump_length As Double

Dim h_0 As Double

Dim tp_1 As Double

velocity = array_source(7, 1)

time = t

wheelbase = array_source(9, 1)

h_0 = array_source(10, 1)

tp_1 = tp1(array_source) 'call fucntion to calculate tp1

bump_length = array_source(19, 1)

z_1 = (1 - (u_tp1(time, tp_1))) * h_0 * _

Sin((WorksheetFunction.Pi * velocity * time) / bump_length)

End Function

Function z_2(array_source, xin, t, dt)


Dim time As Double

Dim wheelbase As Double, bump_length As Double

Dim h_0 As Double

Dim tL As Double, t_p2 As Double



time = t



tL = t_L(array_source)

t_p2 = tp2(array_source)

z_2 = ((u_tL(time, tL)) - (u_tp2(time, t_p2))) * _

h_0 * Sin((WorksheetFunction.Pi * velocity * time) / bump_length)

End Function

Function z_dot1(array_source, xin, t, dt)

Dim t_in As Double, t_p1 As Double 'time constants


Dim time As Double

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Dim wheelbase As Double

Dim h_0 As Double

Dim X As Variant

Dim bump_length As Double

X = xin


t_in = array_source(5, 1)



time = t



z_dot1 = ((WorksheetFunction.Pi * velocity) / bump_length) * (1 - (u_tp1(time, t_p1))) _

* h_0 * Cos((WorksheetFunction.Pi * velocity * time) / bump_length)

End Function

Function z_dot2(array_source, xin, t, dt)

Dim t_in As Double, t_p1 As Double 'time constants


Dim time As Double

Dim wheelbase As Double

Dim h_0 As Double, bump_length As Double

Dim t_p2 As Double, tL As Double

t_in = array_source(5, 1)

t_p1 = array_source(6, 1)


time = t





tL = t_L(array_source)

z_dot2 = ((WorksheetFunction.Pi * velocity) / bump_length) _

* ((u_tL(time, tL)) - (u_tp2(time, t_p2))) * h_0 * _

Cos((WorksheetFunction.Pi * velocity * time) / bump_length)

End Function

Function f_1prime(array_in, xin, t, dt)

Dim X As Variant

X = xin

f_1prime = X(1, 2)

End Function


Dim X As Variant

X = xin

f_3prime = X(1, 4)

End Function




Dim l_1 As Double, l_2 As Double ' lengths

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Dim y_1 As Double, y_2 As Double


Dim mass As Double

Dim X As Variant

X = xin

y_2 = X(1, 2)





l_1 = array_in(11, 1)

l_2 = array_in(12, 1)

y_1 = X(1, 1)

y_3 = X(1, 3)

y_4 = X(1, 4)

mass = array_in(17, 1)

f_2prime = -((k_1 + k_2) * y_1) / mass + (-((c_1 + c_2) * y_2) / mass) _

+ (-(k_1 * l_1 - k_2 * l_2) * y_3) / mass + (-(c_1 * l_1 - c_2 * l_2) * y_4 / mass) _

+ g_1(array_in, xin, t, dt) / mass

End Function




Dim l_1 As Double, l_2 As Double ' lengths



Dim mass As Double, radius As Double

Dim inertia As Double

Dim p1 As Double, p2 As Double

Dim p3 As Double, p4 As Double, p5 As Double

Dim X As Variant

X = xin

y_2 = X(1, 2)

y_4 = X(1, 4)





l_1 = array_in(11, 1)

l_2 = array_in(12, 1)

y_1 = X(1, 1)

y_3 = X(1, 3)

mass = array_in(17, 1)

radius = array_in(18, 1)

inertia = Inertia_moment(mass, radius)

p1 = -(((k_1 * l_1) - (k_2 * l_2)) * y_1) / inertia

p2 = -(((c_1 * l_1) - (c_2 * l_2)) * y_2) / inertia

p3 = -(((k_1 * (l_1) ^ 2) - (k_2 * (l_2) ^ 2)) * y_3) / inertia

p4 = -(((c_1 * (l_1) ^ 2) - (c_2 * (l_2) ^ 2)) * y_4) / inertia

p5 = g_2(array_in, xin, t, dt) / inertia

f_4prime = p1 + p2 + p3 + p4 + p5

End Function

Function tp1(array_in)

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Dim D As Double, v As Double

D = array_in(19, 1) 'length of bump

v = array_in(7, 1)

tp1 = D / v

End Function

Function t_L(array_in)

Dim L As Double

Dim v As Double

L = array_in(9, 1) 'wheelbase

v = array_in(7, 1) 'velocity

t_L = L / v

End Function

Function tp2(array_in)

Dim L As Double

Dim v As Double

Dim D As Double

D = array_in(19, 1) 'length of bump

L = array_in(9, 1) 'wheelbase

v = array_in(7, 1) 'velocity

tp2 = (L + D) / v

End Function

Function rk_4V2(p, xin, t, dt)

Dim X As Variant, n As Double

X = xin 'row array

n = UBound(X, 2) 'number of state variables

K = K_RK4(p, xin, t, dt) 'array function

For i = 1 To n 'state variables

X(1, i) = X(1, i) + (K(1, i) + 2 * K(2, i) + 2 * K(3, i) + K(4, i)) / 6

Next i

rk_4V2 = X

End Function

Function rk_4V3(p, xin, t, dt)

Dim X As Variant, n As Double

Dim x_2 As Variant

Dim B As Variant 'matrix for forcing functions

Dim A_in As Variant 'matrix of input coefficients

Dim Resultant As Variant 'matrix holder for resultant of x-dot minus B

Dim x_out As Variant

ReDim x_out(1, 4)

X = xin 'row array; holds x-dot (gradient function) output matrix

x_2 = p

n = UBound(X, 2) 'number of state variables

K = K_RK4(p, xin, t, dt) 'array function

For i = 1 To n 'state variables

X(1, i) = X(1, i) + (K(1, i) + 2 * K(2, i) + 2 * K(3, i) + K(4, i)) / 6

Next i

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Dim k_1 As Double, k_2 As Double

Dim c_1 As Double, c_2 As Double


Dim wheelbase As Double, height As Double





k_1 = x_2(1, 1)

k_2 = x_2(2, 1)

c_1 = x_2(3, 1)

c_2 = x_2(4, 1)

l_1 = x_2(11, 1)

l_2 = x_2(12, 1)

mass = x_2(17, 1)

radius = x_2(18, 1)


'k1 5000 spring constant, front (1, 1)

'k2 5000 spring constant, rear (1, 2)

'c1 500 damping constant, front (1, 3)

'c2 500 damping constant, rear (1, 4)

't_in 0 ?? (1, 5)

't_p1 15 ?? (1, 6)

'v 6.7 velocity (m/sec) (1, 7)

't 0.5 time (sec) (1, 8)

'L 2 wheelbase (m) (1, 9)

'h 0.1 height (of bump) (1, 10)

'l1 0.8 length (front to cg) (1, 11)

'l2 1.2 length (rear to cg) (1, 12)

'y_1 0 x0 (1, 13)

'y_2 0 x-dot (derivative) (1, 14)

'y_3 0 theta0 (1, 15)

'y_4 0 theta-dot (derivative) (1, 16)

'm 2000 mass (1, 17)

'r0 0.6 radius (1, 18)

'D 1 Length of bump (m) (1, 19)

ReDim B(1 To 1, 1 To 4)

ReDim A_in(1 To 4, 1 To 4)

ReDim Resultant(1 To 4, 1 To 1)

Dim g_one As Double

Dim g_two As Double

g_one = g_1(p, xin, t, dt)

g_two = g_2(p, xin, t, dt)

B(1, 1) = 0

B(1, 2) = g_one / mass

B(1, 3) = 0

B(1, 4) = g_two / inertia

A_in(1, 1) = 0

A_in(1, 2) = 1

A_in(1, 3) = 0

A_in(1, 4) = 1

A_in(2, 1) = -(k_1 + k_2) / mass

A_in(2, 2) = -(c_1 + c_2) / mass

A_in(2, 3) = -(k_1 * l_1 - k_2 * l_2) / mass

A_in(2, 4) = -(c_1 * l_1 - c_2 * l_2) / mass

A_in(3, 1) = 0

A_in(3, 2) = 0

A_in(3, 3) = 0

A_in(3, 4) = 1

A_in(4, 1) = -(k_1 * l_1 - k_2 * l_2) / inertia

A_in(4, 2) = -(c_1 * l_1 - c_2 * l_2) / inertia

A_in(4, 3) = -(k_1 * l_1 ^ 2 + k_2 * l_2 ^ 2) / inertia

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A_in(4, 4) = -(c_1 * l_1 ^ 2 + c_2 * l_2 ^ 2) / inertia

Dim step As Integer

'evaluate x-dot minus b:

For step = 1 To 4

Resultant(step, 1) = X(1, step) - B(1, step)

Next step

X = WorksheetFunction.MMult(WorksheetFunction.MInverse(A_in), Resultant)

Dim step2 As Integer

For step2 = 1 To 4

x_out(1, step2) = X(step2, 1)

Next step2

rk_4V3 = x_out

End Function

Function Inertia_moment(mass, radius)

Inertia_moment = mass * radius ^ 2

End Function

Function K_RK4(p, Xi, t, dt)

Dim X, Xt As Variant

Dim ns As Integer, i As Integer

X = Xi

Xt = Xi 'twicked x

ns = UBound(X, 2) 'number of state variable

Dim K(): ReDim K(1 To 4, 1 To ns)

Dim F(): ReDim F(1 To 1, 1 To ns)

Dim FT(): ReDim FT(1 To 1, 1 To ns)

F = Fbar(p, X, t, dt)

For i = 1 To ns 'Calculate all K1 values

K(1, i) = F(1, i) * dt 'basically same as Euler

Next i

'-------------------------------------------

For i = 1 To ns

Xt(1, i) = X(1, i) + K(1, i) / 2

Next i

FT = Fbar(p, Xt, t + dt / 2, dt)


K(2, i) = FT(1, i) * dt 'basically same as Euler

Next i

'--------------------------

For i = 1 To ns

Xt(1, i) = X(1, i) + K(2, i) / 2

Next i

FT = Fbar(p, Xt, t + dt / 2, dt)



Next i

'---------------------------------------------

For i = 1 To ns

Xt(1, i) = X(1, i) + K(3, i)

Next i

FT = Fbar(p, Xt, t + dt, dt)



Next i

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K_RK4 = K

End Function

Function Fbar(p, Xi, t, dt)

Dim X As Variant

Dim ns As Integer

X = Xi

ns = UBound(X, 2) 'number of state variables

Dim F(): ReDim F(1 To 1, 1 To ns)

F(1, 1) = f_1prime(p, X, t, dt)

F(1, 2) = f_2prime(p, X, t, dt)

F(1, 3) = f_3prime(p, X, t, dt)

F(1, 4) = f_4prime(p, X, t, dt)

Fbar = F

End Function

Function Euler(p, Xi, t, dt)

Dim ns As Integer, i As Integer

Dim x_2 As Variant

Dim X As Variant 'state vector

Dim F As Variant 'state gradient vector-column topology

Dim x_out As Variant

Dim D_Identity As Variant

ReDim D_Identity(1 To 4, 1 To 4)

ReDim x_out(1, 4)

X = Xi 'Initial State - assume row topology

ns = UBound(X, 2) 'number of state variables

x_2 = p

F = Fbar(p, X, t, dt)

For i = 1 To ns 'state variables

X(1, i) = X(1, i) + F(1, i) * dt 'Eulers Eqn

Next i

'X(1, 1) = f_1prime(p, Xi, t, dt)

'X(1, 2) = f_2prime(p, Xi, t, dt)

'X(1, 3) = f_3prime(p, Xi, t, dt)

'X(1, 4) = f_4prime(p, Xi, t, dt)

Dim k_1 As Double, k_2 As Double

Dim c_1 As Double, c_2 As Double


Dim wheelbase As Double, height As Double





k_1 = x_2(1, 1)

k_2 = x_2(2, 1)

c_1 = x_2(3, 1)

c_2 = x_2(4, 1)

l_1 = x_2(11, 1)

l_2 = x_2(12, 1)

mass = x_2(17, 1)

radius = x_2(18, 1)


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ReDim B(1 To 1, 1 To 4)

ReDim A_in(1 To 4, 1 To 4)

ReDim Resultant(1 To 4, 1 To 1)

Dim g_one As Double

Dim g_two As Double

g_one = g_1(p, xin, t, dt)

g_two = g_2(p, xin, t, dt)

B(1, 1) = 0

B(1, 2) = g_one / mass

B(1, 3) = 0

B(1, 4) = g_two / inertia

A_in(1, 1) = 0

A_in(1, 2) = 1

A_in(1, 3) = 0

A_in(1, 4) = 1

A_in(2, 1) = -(k_1 + k_2) / mass

A_in(2, 2) = -(c_1 + c_2) / mass

A_in(2, 3) = -(k_1 * l_1 - k_2 * l_2) / mass

A_in(2, 4) = -(c_1 * l_1 - c_2 * l_2) / mass

A_in(3, 1) = 0

A_in(3, 2) = 0

A_in(3, 3) = 0

A_in(3, 4) = 1

A_in(4, 1) = -(k_1 * l_1 - k_2 * l_2) / inertia

A_in(4, 2) = -(c_1 * l_1 - c_2 * l_2) / inertia

A_in(4, 3) = -(k_1 * l_1 ^ 2 + k_2 * l_2 ^ 2) / inertia

A_in(4, 4) = -(c_1 * l_1 ^ 2 + c_2 * l_2 ^ 2) / inertia

D_Identity(1, 1) = 1
















Dim step As Integer

'evaluate x-dot minus b:

For step = 1 To 4

Resultant(step, 1) = X(1, step) - B(1, step)

Next step

X = WorksheetFunction.MMult(WorksheetFunction.MInverse(A_in), Resultant)

Dim step2 As Integer

For step2 = 1 To 4

x_out(1, step2) = X(step2, 1)

Next step2

Euler = x_out

End Function

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Car Modeling and Simulation