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
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)
'
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:
(
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: z
1, z
2.
• Second assumption: We assume that the car’s speed is constant (acceleration is
zero) at this period.
Speed
Car
V
V
D
t
D
d
h
z
.
sin
.
d Vt p:
1 . 0 1=
→
=
=π
=
,
0
.
sin
.
0 1D
d
h
z
π
1 1
tp
t
if
tp
t
if
≥
→
<
→
=
0
.
.
sin
.
0
0 2D
t
V
h
z
π
2 2tp
t
if
tp
t
t
if
t
t
if
l l>
→
<
<
→
<
→
Note:
V
D
L
tp
V
L
t
l+
=
→
=
2Top graph: Front wheel
h: height of the bump
d: horizontal coordinate
D: length of the bump
L: distance between the front and back tires
Other form of z
1
, z
2
function:
Introduce
=
1
0
)
(t
U
cC
t
C
t
>
≤
(
)
[
−
]
=
−
=
D
t
V
h
t
U
t
U
z
D
t
V
h
t
U
z
tp tl tp.
.
sin
.
.
)
(
)
(
.
.
sin
.
.
)
(
1
0 0 2 0 0 1 2 1π
π
Check,
0
.
.
sin
.
).
1
1
(
.
.
sin
.
.
.
sin
.
).
0
1
(
0
.
.
sin
.
).
0
0
(
.
.
sin
.
).
0
1
(
0 0 2 2 0 0 0 0 2 2 0 0 2 0 0 1 1=
−
=
→
>
=
−
→
≤
≤
=
−
=
→
<
−
=
→
<
D
t
V
h
z
tp
t
D
t
V
h
D
t
V
h
z
tp
t
t
D
t
V
h
z
t
t
D
t
V
h
z
tp
t
L Lπ
π
π
π
π
Differential Equation of Motion 2DOF (linear)
Assume:
θ
θ
2 1 2 2l
x
z
l
x
z
+
<
−
<
=
m
. x
..F
x .. 1 1 2 2Fc
Fc
Fs
m
x
Fs
−
−
−
=
−
)
(
)
(
2 1 1 1 2 1 2 2z
l
x
k
Fs
z
l
x
k
Fs
−
−
=
−
−
=
θ
θ
)
(
)
(
2 . . 1 . 1 1 2 . . 2 . 2 2z
l
x
c
Fc
z
l
x
c
Fc
−
−
=
−
−
=
θ
θ
.. 1 .. 1 .. 1 1 . 1 1 1 1 1 1 2 . 2 . 2 2 . 2 2 2 2 2 2x
k
l
k
z
c
x
c
l
c
z
k
x
k
l
k
z
c
x
c
l
c
z
m
x
k
+
+
−
+
+
−
−
+
−
−
+
=
−
θ
θ
θ
θ
2 2 1 1 2 . 2 1 . 1 2 2 1 1 2 1 . 2 2 1 1 . 2 1 ..)
(
)
(
)
(
)
(
c
c
x
c
l
c
l
k
k
x
k
l
k
l
c
z
c
z
k
z
k
z
x
m
+
+
+
−
θ
+
+
+
−
θ
=
+
+
+
)
=
..θ
I
Mc
G .. 1 1 1 1 2 2 2 2.
l
Fc
l
Fc
.
l
Fs
l
I
θ
Fs
+
−
−
=
.. 1 1 1 1 1 . . 1 . 1 1 2 . . 2 . 2 2 2 2 2 2l
(
x
l
θ
z
)
c
l
(
x
l
θ
z
)
c
l
(
x
l
θ
z
)
k
l
(
x
l
θ
z
)
I
θ
k
−
−
+
−
−
−
+
−
−
+
−
=
z
l
k
z
l
k
z
l
c
z
l
c
x
l
k
l
k
l
k
l
k
x
l
c
l
c
l
c
l
c
I
2 11 1 2 2 . 2 2 1 . 1 1 2 2 1 1 2 1 1 2 2 2 . 2 2 1 1 . 2 1 1 2 2 2 ..)
(
)
(
)
(
)
(
+
+
−
+
+
+
−
=
−
+
−
+
θ
θ
θ
If we put 1, 2 in matrix format;
+
−
−
+
+
+
−
−
+
θ
θ
θ
x
l
k
l
k
l
k
l
k
l
k
l
k
k
k
x
l
c
l
c
l
c
l
c
l
c
l
c
c
x
I
m
.
.
.
0
0
2
1
1
2
2
2
2
2
1
1
2
2
1
1
2
1
.
.
2
1
1
2
2
2
2
2
1
1
2
2
1
1
1
..
..
−
+
−
+
+
+
=
2
2
2
1
1
1
2
.
2
2
1
.
1
1
2
2
1
1
2
.
2
1
.
1
z
l
k
z
l
k
z
l
c
z
l
c
z
k
z
k
z
c
z
c
*
Calculation of Natural Frequency and Mode shapes
Assuming Reasonable values for all parameter, we have:
2
.r
m
I
=
r : Radius gyration of the car
Reasonable approximate values for a Sedan car;
2 2
6
.
0 m
r
=
m
=
1000
kg
c
1=
c
2=
500
N
.
s
m
m
N
k
k
1=
2=
5000
l
1=
0
.
8
m
l
2=
1
.
2
m
Reasonable approximate values for a SUV;
2 2
65
.
0
m
r
=
m
=
3000
kg
c
1=
c
2=
1500
N
.
s
m
m
N
k
k
1=
2=
1500
l
1=
1
m
l
2=
1
.
5
m
Solved for a
bump
with
h
=
0
.
1
m
,
D
=
1
.
5
m
Eigen Value Problem and Mode Shapes
For our model we assume c
1= c
2= c and k
1= k
2= k
=
+
−
−
+
+
−
−
+
→
)
(
)
(
.
)
(
)
(
)
(
2
.
)
(
)
(
)
(
2
.
0
0
2 1 2 1 2 1 2 1 2 1 . . 2 1 2 1 2 1 2 1 .. .. 3 .t
g
t
g
x
l
l
k
l
l
k
l
l
k
k
x
l
l
c
l
l
c
l
l
c
c
x
I
m
eqθ
θ
θ
Where:
[
]
[
]
[
1
(
)
]
.(
.
sin
,
.
cos
)
.
[
(
)
(
)
]
.(
.
sin
,
.
cos
)
.
)
(
)
cos
.
,
sin
.
.(
)
(
)
(
)
cos
.
,
sin
.
.(
)
(
1
)
(
2 1 2 1 2 0 1 0 2 0 0 1t
c
t
k
t
u
t
u
l
h
t
c
t
k
t
u
l
h
t
g
t
c
t
k
t
u
t
u
h
t
c
t
k
t
u
h
t
g
f f f tp tL f f f tp f f f tp tL f f f tpω
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
+
−
+
+
−
=
+
−
+
+
−
=
Note:
D
V
f.
π
ω
=
+
Eigen Value Problem:
[ ]
[ ]
(
k
−
ω
2.
m
)
=
0
0
.
)
(
)
(
)
(
2
det
2 2 2 2 1 2 1 2 1 2=
+
−
−
−
I
l
l
k
l
l
k
l
l
k
k
ω
ω
0
.
)
(
.
)
(
2
1 2 2 2 2 2 2 2 1 4=
+
+
+
+
−
I
M
l
l
k
l
l
I
k
m
k
ω
ω
I
m
l
l
k
l
l
I
k
m
k
l
l
I
k
m
k
.
)
(
4
)
(
2
)
(
2
2 2 2 1 2 2 2 2 1 2 2 2 1 2=
+
+
±
+
+
−
+
ω
For a sedan car we have:
kg
m
=
1000
,
.r
2m
I
=
and
20
.
6
600
2Nm
I
r
=
=
8
.
0
1=
l
l
2=
1
.
2
,
k
1=
k
2=
5000
N
m
c
1=
c
2=
500
N
.
s
m
If we plug in eq.(5)
0
600
1000
)
2
.
1
8
.
0
.(
)
5000
(
)
2
.
1
8
.
0
(
600
5000
1000
10000
2 2 2 2 2 4=
⋅
+
+
+
+
−
ω
ϖ
0
677
.
166
334
.
27
2 4=
+
−
ϖ
ω
\
182
.
9
2 1=
ω
ω
22=
18
.
151
)
sec
(
03
.
3
1=
rad
ω
ω
2=
4
.
26
(
rad
sec
)
Back sub
1in (4)
)
.
(
409
.
0
1
8
.
4890
2000
2000
818
)
600
).(
182
.
9
(
)
2
.
1
8
.
0
.(
5000
)
2
.
1
8
.
0
.(
5000
)
2
.
1
8
.
0
.(
5000
)
1000
).(
182
.
9
(
10000
.
)
(
)
(
)
(
2
1 2 2 2 1 2 2 2 1 2 1 2 1 2 1Mode
First
I
l
l
k
l
l
k
l
l
k
m
k
→
=
−
−
=
−
+
−
−
−
=
−
+
−
−
−
ϕ
ω
ω
Sub
2in (4)
)
.
(
075
.
4
1
6
.
490
2000
2000
8151
)
600
).(
151
.
18
(
)
2
.
1
8
.
0
.(
5000
)
2
.
1
8
.
0
.(
5000
)
2
.
1
8
.
0
.(
5000
)
1000
).(
151
.
18
(
10000
2 2 2Mode
Second
→
−
−
−
−
−
=
−
+
−
−
−
ϕ
∴
Approximate Mode Shapes and Natural Frequencies
For Sedan Cars are,
−
=
=
=
=
075
.
4
1
sec
26
.
4
409
.
0
1
sec
03
.
3
2 2 1 1ϕ
ω
ϕ
ω
rad
rad
For an SUV we have
kg
m
=
3000
2/ r
m
I
=
and
r
=
0
.
64
→
I
=
1920
N
/
m
2m
N
k
k
1=
2=
15000
l
1=
1
l
2=
1
.
5
m
N
c
c
1=
2=
1500
Plug the numbers in (5),
0
)
1920
).(
3000
(
)
5
.
1
1
.(
)
15000
(
.
5
.
1
1
(
1920
15000
3000
30000
2 2 2 2 2 4−
+
+
ω
+
+
=
ω
0
14
.
244
39
.
35
2 4−
ω
+
=
ω
39
.
9
2 1=
→
ω
,
ω
22=
26
sec
064
.
3
1=
rad
ω
,
ω
2=
5
.
1
rad
sec
Back sub
ω
1in (4),
−
−
=
−
+
−
−
−
2
.
30721
7500
7500
1830
)
1920
).(
39
.
9
(
)
5
.
1
1
.(
15000
)
5
.
1
1
.(
15000
)
5
.
1
1
.(
15000
)
3000
).(
39
.
9
(
30000
2=
→
244
.
0
1
1ϕ
Back sub
2in (4),
−
−
−
−
=
−
+
−
−
−
1170
7500
7500
48000
)
1920
).(
26
(
)
5
.
1
1
.(
15000
)
5
.
1
1
.(
15000
)
5
.
1
1
.(
15000
)
3000
).(
26
(
30000
2−
=
→
4
.
6
1
2ϕ
∴
Approximate Mode Shapes and Natural Frequencies
For an SUV is:
−
=
=
=
=
4
.
6
1
sec
1
.
5
244
.
0
1
sec
06
.
3
2 2 1 1ϕ
ω
ϕ
ω
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 1 2 1 2 1 2 1 2 1 2 1 2 12
)
.(
)
.(
)
.(
.
2
l
l
l
l
l
l
k
l
l
k
l
l
k
l
l
k
k
k
'
[ ]
+
−
−
=
+
−
−
=
2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 12
.
)
.(
)
.(
)
.(
.
2
l
l
l
l
l
l
C
l
l
C
l
l
C
l
l
C
C
C
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
.
4
409
.
0
1
1
ϕ
,
=
600
0
0
1000
M
[ ] [ ][ ]
=
=
10963
0
0
1100
.
.
0
0
2 1ϕ
ϕ
m
m
m
Tkg
m
1=
1100
,
m
2=
10963
kg
Modal Damping: (sedan)
[ ] [ ][ ] [ ]
−
−
=
=
1090
200
200
1000
,
.
.
0
0
2 1C
C
C
C
Tϕ
ϕ
=
19900
0
0
1010
0
0
2 1C
C
m
s
N
C
1=
1010
.
,
C
2=
19900
N
.
s
m
Modal Stiffness: (sedan)
[ ] [ ][ ] [ ]
−
−
=
=
10400
2000
2000
10000
,
.
.
0
0
2 1k
m
k
k
Tϕ
ϕ
=
199000
0
0
10100
0
0
2 1k
k
m
N
k
1=
10100
.
k
2=
199000
N
.
m
∴
Our uncoupled System is,
[ ]
=
2 1.
η
η
ϕ
θ
x
(
=
+
+
=
+
+
)
_
(mod
.
.
)
_
(mod
.
.
2 2 2 2 . 2 2 .. 2 1 1 1 . 1 1 .. 1 1force
al
F
k
C
m
force
al
F
k
C
m
η
η
η
η
η
η
m
1, c
1, k
1, m
2, k
2, c
2, are already calculated. We should find only F
1and F
2which are
model forces.
Modal Damping: (For SUV Car)
[ ]
−
−
=
+
−
−
=
4875
750
750
3000
)
.(
)
.(
)
.(
.
2
'
2 1 2 1 2 1 2 1l
l
C
l
l
C
l
l
C
C
C
[ ] [ ][ ]
=
=
21280
0
0
2920
'
.
'
.
'
0
0
' 2 ' 1ϕ
C
ϕ
C
C
Tm
s
N
C
1'=
2920
.
,
C
2'=
212280
N
.
s
m
Modal Stiffness: (For SUV Car)
[ ]
−
−
=
+
−
−
=
48750
7500
7500
30000
)
.(
)
.(
)
.(
.
2
'
2 1 2 1 2 1 2 1l
l
k
l
l
k
l
l
k
k
k
=
212800
0
0
29200
0
0
' 2 ' 1k
k
m
N
k
1'=
29200
.
,
k
2'=
2122800
N
.
m
∴
Our uncoupled system is:
[ ]
=
2 1.
η
η
ϕ
ϕ
x
=
+
+
=
+
+
)
_
(mod
.
.
)
_
(mod
.
.
' 2 ' 2 . ' 2 .. ' 2 ' 1 ' 1 . ' 1 .. ' 1force
al
F
k
C
m
force
al
F
k
C
m
η
η
η
η
η
η
Calculation of Modal forces:
[ ]
(6)
(t)
g
(t)
g
F
F
2 1 T 2 1=
Finding g
1and g
2 f tP tL tP tL tP tAD
v
D
vt
t
U
t
U
D
v
z
D
vt
t
U
t
U
z
D
vt
t
U
D
v
z
D
vt
t
U
z
z
k
z
k
z
c
z
c
t
g
t
g
t
g
ω
π
π
π
π
π
π
π
=
−
=
→
−
=
−
=
→
−
=
+
+
+
=
cos
h
)]
(
)
(
[
sin
h
)]
(
)
(
[
cos
h
)]
(
1
[
sin
h
)]
(
1
[
)
(
)
(
and
)
(
Calculate
o 2 o 2 o 1 o 1 2 2 1 1 2 2 1 1 1 2 1 2 2 1V
D
L
V
L
V
D
x
By
t
w
w
c
t
w
t
U
t
U
l
t
w
w
c
t
w
k
t
U
l
t
g
z
l
k
z
l
c
z
l
k
z
l
c
t
g
The
t
w
w
c
t
w
t
U
t
U
t
w
w
c
t
w
k
t
U
t
g
t
w
t
U
t
U
t
w
t
U
k
t
w
t
U
t
U
w
c
t
w
t
U
w
c
t
g
f f f tP tL f f f tP f f f tP tL f f f tP f tP tL f tP i f tP tL f f tP f+
=
=
=
+
−
−
+
−
=
−
−
+
=
+
−
+
+
−
=
−
+
−
+
−
+
−
=
2 1 2 1 2 1 2 1 2 1 P L P 2 2 2 o 1 1 1 o 2 2 2 2 2 2 2 1 1 1 1 1 1 2 2 2 o 1 1 o 1 o 2 o o 2 o 1 1t
,
t
,
t
:
codes
in
that
Note
,
,
for x,
solution
numerical
a
have
will
we
Method,
Kutta
Runge
in
Codes
VBA
using
)
cos
sin
k
)](
(
)
(
[
h
)
cos
sin
)](
(
1
[
h
)
(
)
(
for
method
same
)
cos
sin
k
)](
(
)
(
[
h
)
cos
sin
)](
(
1
[
h
)
(
sin
sin
)]
(
)
(
[
h
k
sin
)]
(
1
[
h
cos
)]
(
)
(
[
h
cos
)]
(
1
[
h
)
(
θ
θ
500
c
5000,
K
,
m
2
2
.
1
8
.
0
L
L
and
bumb)
the
of
length
and
(height
1.5
D
m
1
.
0
h
with
)
(6
in
put them
and
)
(
and
)
(
calculate
should
we
So
2 1 o 2 1
=
=
=
+
=
+
=
=
=
L
t
g
t
g
sec
0.52
t
sec
522
.
0
t
sec
3
.
0
t
sec
298
.
0
7
.
6
2
t
sec
22
.
0
t
sec
224
.
0
7
.
6
5
.
1
t
rad/sec
03
.
14
5
.
1
7
.
6
m/sec
6.7
mph
15
V
1)
2 2 1 1 P P L L P P f=
≈
=
+
=
=
≈
=
=
=
=
≈
=
=
=
=
×
=
=
→
=
=
v
D
L
L
v
v
D
D
v
π
π
ω
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
z
1
,
z
2
(
h
e
ig
h
t,
m
e
te
rs
)
z1
z2
)
sec
0.29
t
sec
292
.
0
12
5
.
3
t
sec
17
.
0
t
sec
167
.
0
12
2
t
sec
12
.
0
t
sec
125
.
0
12
5
.
1
t
rad/sec
13
.
25
5
.
1
12
w
m/sec
12
mph
27
v
)
2
)]
03
.
14
cos(
8
.
841
)
03
.
14
sin(
600
)][
(
)
(
[
)]
03
.
14
cos(
2
.
562
)
03
.
14
sin(
400
)][
(
1
[
)
(
)]
03
.
14
cos(
5
.
701
)
03
.
14
sin(
500
)][
(
)
(
[
)]
03
.
14
cos(
5
.
701
)
03
.
14
sin(
500
)][
(
1
[
)
(
formula,
into
numbers
e
Insert th
2 2 1 1 P P L L P P f 52 . 0 3 . 0 22 . 0 2 52 . 0 3 . 0 22 . 0 1=
≈
=
=
+
=
=
≈
=
=
=
=
≈
=
=
=
=
×
=
=
→
=
=
+
−
+
+
−
=
+
−
+
+
−
=
v
D
L
L
v
v
D
D
v
t
t
t
U
t
U
t
t
t
U
t
g
t
t
t
U
t
U
t
t
t
U
t
g
π
π
With this velocity the graphs for
g
1(
t
)
&
g
2(
t
)
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
g
1
,
g
2
g1
g2
*
sec
0.29
t
sec
292
.
0
12
5
.
3
t
sec
17
.
0
t
sec
167
.
0
12
2
t
sec
12
.
0
t
sec
125
.
0
12
5
.
1
t
rad/sec
13
.
25
5
.
1
12
w
m/sec
12
mph
27
v
)
2
2 2 1 1 P P L L P P f=
≈
=
=
+
=
=
≈
=
=
=
=
≈
=
=
=
=
×
=
=
→
=
=
v
D
L
L
v
v
D
D
v
π
π
)]
44
cos(
2640
)
44
sin(
600
)][
(
)
(
[
)]
44
cos(
1760
)
44
sin(
400
)][
(
1
[
)
(
)]
44
cos(
2200
)
44
sin(
500
)][
(
)
(
[
)]
44
cos(
2200
)
44
sin(
500
)][
(
1
[
)
(
sec
0.16
t
sec,
09
.
0
t
sec,
07
.
0
t
rad/sec
44
w
m/sec
21
mph
47
v
)
3
)]
15
.
25
cos(
8
.
1507
)
13
.
25
sin(
600
)][
(
)
(
[
)]
13
.
25
cos(
2
.
1005
)
13
.
25
sin(
400
)][
(
1
[
)
(
)]
13
.
25
cos(
5
.
1256
)
13
.
25
sin(
500
)][
(
)
(
[
)]
13
.
25
cos(
5
.
1256
)
13
.
25
sin(
500
)][
(
1
[
)
(
formula,
into
number
he
Insert t
16 . 0 09 . 0 07 . 0 2 16 . 0 09 . 0 07 . 0 1 P L P f 29 . 0 17 . 0 12 . 0 2 29 . 0 17 . 0 12 . 0 1 2 1t
t
t
U
t
U
t
t
t
U
t
g
t
t
t
U
t
U
t
t
t
U
t
g
D
v
t
t
t
U
t
U
t
t
t
U
t
g
t
t
t
U
t
U
t
t
t
U
t
g
+
−
+
+
−
=
+
−
+
+
−
=
=
=
=
=
=
→
=
=
+
−
+
+
−
=
+
−
+
+
−
=
π
)
(
075
.
4
)
(
199000
19900
10963
)
(
409
.
0
)
(
10100
1010
1100
Equation"
Uncoupled
"
have
we
,
cases
all
For
(8)
)
(
075
.
4
)
(
and
)
(
409
.
0
)
(
)
(
)
(
4.075
1
0.409
1
2 1 2 2 2 2 1 1 1 1 2 1 2 2 1 1 2 1 2 1
t
g
t
g
t
g
t
g
t
g
t
g
F
t
g
t
g
F
t
g
t
g
F
F
−
=
+
+
+
=
+
+
−
=
+
=
=
′
η
η
η
η
η
η
+
General solution
[ ]
213
.
0
2
,
10963
m
,
26
.
4
151
.
0
2
,
1100
m
,
03
.
3
2
.
1
)
(
sin
)
(
1
0
B
A
have
we
zero,
is
condition
initial
that
assuming
]
[
B
,
]
[
A
1
w
(7)
)
(
sin
)
(
1
]
sin
)
(
cos
[
2 2 2 2 2 2 1 1 1 1 1 1 0 k k 0 0 k 0 0 k 2 dk 0≈
=
=
=
≈
=
=
=
=
−
′
×
=
=
=
×
×
×
Φ
=
×
×
Φ
=
−
=
−
′
+
+
+
=
−w
m
c
w
w
m
c
w
k
d
t
w
e
F
m
w
By
m
w
x
m
m
x
m
w
Where
d
t
w
e
F
w
m
w
w
A
w
B
t
w
A
e
dk t A w k k dk k k dk T k k T k k dk t A w k dk k dk dk k w k k dk k t w k k w k k w k k kξ
ξ
τ
τ
τ
η
ϑ
ϑ
ξ
τ
τ
τ
ξ
η
ξ ξ ξrad/sec
16
.
4
-1
rad/sec
99
.
2
-1
2 2 2 2 2 1 1 1
=
=
=
=
ξ
ξ
w
w
w
w
and
d d[ ]
2 1 2 1 2 1 2 1 0 9 . 0 2 2 0 46 . 0 1 14.075
409
.
0
4.075
0.409
1
1
advised
are
solutions
numerical
the
so
evluated,
be
to
d
complecate
so
is
it
MATLAB
in
even
but
integral
those
calculated
can
we
(t)
F
(t),
F
different
if
caused
which
speed
different
with
(9)
in
(8)
from
)
(
16
.
4
sin
)
(
45606
1
(9)
)
(
99
.
2
sin
)
(
328
1
η
η
θ
η
η
η
η
η
η
θ
η
η
−
=
+
=
=
Φ
=
−
=
−
=
− −x
x
Substitude
dz
z
t
e
z
F
dz
z
t
e
z
F
t z t zDifferential Equation of Motion Using Lagrange Eq
2 2 1 1 2 2 1 1 2 2 1 1 2 1 2 2 1 1 2 1 1 1 1 2 2 1 1 2 2 2 2 2 2 1 1 1 1 1 2 2 2 1 1 1 2 2 2 1 1 1 2 2 2 2 1 1 1 1 1 1 2 2 1 1 2 2 2 1 2 1 2 1 1 2 2 2 2 1 1 2 2 2 2 0)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
0
t
:
.
t
)
2
t
)
1
)
(
2
1
)
(
2
1
)
(
2
1
)
(
2
1
2
1
T
z
k
z
k
z
c
z
c
l
k
l
k
x
k
k
l
c
l
c
x
c
c
x
m
Q
l
x
k
l
x
k
l
x
c
l
x
c
x
m
z
k
z
c
z
k
z
c
Q
z
l
k
z
l
k
z
l
c
z
l
c
x
z
k
z
c
z
k
z
c
w
l
x
k
l
x
k
x
V
l
x
c
l
x
c
x
R
x
T
x
m
x
T
x
m
x
T
equation
x
Q
V
R
T
T
Q
x
V
x
R
x
T
x
T
Q
x
Q
w
l
x
c
l
x
c
R
l
x
k
l
x
k
V
I
x
m
+
+
+
=
−
+
+
+
−
+
+
+
=
+
+
−
+
+
+
−
+
+
+
+
=
−
+
−
+
∂
+
+
+
=
∂
+
+
−
=
∂
∂
+
+
−
=
∂
∂
=
∂
∂
=
∂
∂
∂
∂
→
=
∂
∂
=
∂
∂
+
∂
∂
+
∂
∂
−
∂
∂
∂
∂
=
∂
∂
+
∂
∂
+
∂
∂
−
∂
∂
∂
∂
∂
+
∂
=
∂
+
+
−
=
+
+
−
=
+
=
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
2 2 2 1 1 1 2 2 2 1 1 1 2 2 2 2 1 1 2 2 1 1 2 1 1 2 2 2 2 2 1 1 2 2 2 1 1 1 2 2 2 1 1 1 1 1 1 1 2 2 2 1 1 1 2 2 2 2 2 1 1 2 2 1 1 2 2 1 1 2 1 2 2 1 1 2 1 1 1 1 2 2 1 1 2 2
)
)
(
)
(
)
(
)
(
)
(
)
(
)
(
0
t
:
.
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
z
l
k
z
l
k
z
l
c
z
l
c
l
k
l
ck
x
l
k
l
k
l
c
l
c
x
l
c
l
c
I
z
l
k
z
l
k
z
l
c
z
l
c
Q
and
l
x
l
k
l
x
l
k
V
l
x
l
c
l
x
l
c
R
T
I
T
I
T
equation
z
k
z
k
z
c
z
c
l
k
l
k
x
k
k
l
c
l
c
x
c
c
x
m
Q
l
x
k
l
x
k
l
x
c
l
x
c
x
m
−
+
−
=
+
+
−
+
+
+
−
+
+
−
+
−
=
+
+
−
−
=
∂
∂
+
+
−
=
∂
∂
=
∂
∂
=
∂
∂
∂
∂
→
=
∂
∂
+
+
+
=
−
+
+
+
−
+
+
+
=
+
+
−
+
+
+
−
+
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
Put them in matrix format:
+
−
−
+
+
+
−
−
+
θ
θ
θ
x
l
k
l
k
l
k
l
k
l
k
l
k
k
k
x
l
c
l
c
l
c
l
c
l
c
l
c
c
x
I
m
.
.
.
0
0
2 1 1 2 2 2 2 2 1 1 2 2 1 1 2 1 . . 2 1 1 2 2 2 2 2 1 1 2 2 1 1 1 .. ..−
+
−
+
+
+
=
2 2 2 1 1 1 2 . 2 2 1 . 1 1 2 2 1 1 2 . 2 1 . 1z
l
k
z
l
k
z
l
c
z
l
c
z
k
z
k
z
c
z
c
Differential Equation of Motion 4DOF (linear)
2 2 1 1 1 2 3 2 1 2 2 1 2 4 2 2 3 3 2 2 1 1 1 2 3 2 1 2 2 2 2 2 3 3 2 0 2 1)
(
2
1
)
(
2
1
)
(
2
1
)
(
2
1
)
(
2
1
)
(
2
1
2
1
2
1
2
1
T
x
l
x
c
x
l
x
c
R
z
x
k
z
x
k
x
l
x
k
x
l
x
k
V
x
m
x
m
I
x
m
−
+
+
−
−
=
−
+
−
+
−
+
+
−
−
=
+
+
+
=
θ
θ
θ
θ
θ
0
t
)
4
0
t
)
3
0
t
)
2
0
t
)
1
,
any
have
t
don'
we
so
energy,
potential
in the
,
bring
we
case
this
3 3 3 3 2 2 2 2 1 1 1 1 2 1 2 1=
∂
∂
+
∂
∂
+
∂
∂
−
∂
∂
∂
∂
=
∂
∂
+
∂
∂
+
∂
∂
−
∂
∂
∂
∂
=
∂
∂
+
∂
∂
+
∂
∂
−
∂
∂
∂
∂
=
∂
∂
+
∂
∂
+
∂
∂
−
∂
∂
∂
∂
x
V
x
R
x
T
x
T
x
V
x
R
x
T
x
T
V
R
T
T
x
V
x
R
x
T
x
T
Q
Q
z
z
In
θ
θ
θ
θ
'
0
)
(
0
)
(
)
)(
(
)
)(
(
)
4
0
)
(
0
)
(
)
)(
(
)
)(
(
)
3
0
)
(
)
(
)
(
)
(
0
)
(
)
)(
(
)
(
)
)(
(
)
2
0
)
(
)
(
)
(
)
(
0
)
(
)
(
)
(
)
(
)
1
2 3 2 2 1 2 3 3 2 2 2 1 2 3 2 3 3 2 3 3 3 2 1 2 3 2 1 2 3 3 1 4 1 1 1 1 2 4 1 1 1 1 1 2 1 2 2 1 2 4 2 1 1 1 2 1 1 2 2 2 3 2 2 2 1 1 1 2 2 1 1 2 2 2 2 1 1 3 2 2 2 1 1 1 2 2 1 1 2 1 1 2 2 2 2 1 1 1 1 3 2 1 2 2 2 1 1 1 1 3 2 1 2 2 3 2 2 1 2 2 1 1 1 2 1 3 2 2 1 2 2 1 1 1 2 1 1 1 2 1 1 1 3 2 1 2 2 1 1 1 3 2 1 2 1 1=
−
+
−
+
+
+
−
+
=
−
+
−
−
−
+
−
−
−
+
=
−
−
−
+
+
−
−
+
=
−
+
−
+
−
+
−
+
−
+
=
+
−
−
+
+
+
+
−
−
+
+
+
=
−
+
+
−
−
−
+
−
+
+
−
−
−
+
=
−
−
−
+
+
+
−
−
−
+
+
+
=
−
+
+
−
−
+
−
+
+
−
−
+
z
k
l
k
x
k
x
k
k
l
c
x
c
x
c
x
m
z
x
k
x
l
x
k
x
l
x
c
x
m
z
k
l
k
x
k
x
k
k
l
c
x
c
x
c
x
m
z
x
k
x
l
x
k
x
l
x
c
x
m
x
l
k
x
l
k
x
l
k
l
k
l
k
l
k
x
l
c
x
l
c
x
l
c
l
c
l
c
l
c
I
x
l
x
l
k
x
l
x
l
k
x
l
x
l
c
x
l
x
l
c
I
x
k
x
k
l
k
l
k
x
k
k
x
c
x
c
l
c
l
c
x
c
c
x
m
x
l
x
k
x
l
x
k
x
l
x
c
x
l
x
c
x
m
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
Put them in matrix format:
s.
variable
state
eight
have
we
case
In this
:
0
0
x
k
0
k
-0
k
c
0
c
-0
c
m
0
0
0
0
m
0
0
0
0
I
0
0
0
0
2 3 1 4 3 2 1 2 2 2 2 1 1 1 1 2 2 1 1 2 2 2 2 1 1 2 2 1 1 2 1 2 2 1 1 2 1 3 2 1 2 2 2 2 1 1 1 1 2 2 1 1 2 2 2 2 1 1 2 2 1 1 2 1 2 2 1 1 2 1 3 2 1 3 2 1
Note
z
k
z
k
x
x
l
k
k
l
k
l
k
l
k
l
k
l
k
l
k
l
k
k
k
l
k
l
k
k
k
x
x
x
l
c
l
c
c
l
c
l
c
l
c
l
c
l
c
l
c
c
c
l
c
l
c
c
c
x
x
x
m
=
+
−
−
−
−
+
+
−
+
−
−
−
−
+
+
θ
θ
θ
(
Differential Equation of Motion 2DOF (Non linear)
2 1 1 2 2 2 2 1 1 2 2 2 2 0