Biomedical Engineering Reference
In-Depth Information
Step 3:
In a similar manner, we transform the ankle moments from the global
to the anatomical axes system:
⎡
⎤
⎡
⎤
⎡
⎤
M
xd
M
yd
M
zd
0
.
8955
−
0
.
4363
0
.
0875
M
XD
M
YD
M
ZD
⎣
⎦
=
⎣
⎦
⎣
⎦
0
.
4443
0
.
8877
−
0
.
121
−
0
.
025
0
.
1473
0
.
9888
⎡
⎤
⎡
⎤
⎡
⎤
0
.
8955
−
0
.
4363
0
.
0875
14
.
29
2
.
11
2
.
79
20
.
79
⎣
⎦
⎣
⎦
=
⎣
⎦
=
0
.
4443
0
.
8877
−
0
.
121
−
0
.
025
0
.
1473
0
.
9888
−
103
.
85
−
102
.
73
Using Equation (7.7
b
), we calculate the angular velocities and acceler-
ations required for the solution of Euler's kinetic Equations (7.9). As
previously calculated for frame 6,
c
2
=
0
.
9997,
c
3
=
0
.
8958,
s
2
=−
0
.
025,
s
3
=−
0
.
4444.
⎡
⎤
⎡
⎤
⎡
⎤
⎡
⎤
⎡
⎤
θ
1
θ
2
θ
3
ω
x
ω
y
ω
z
c
2
c
3
s
3
0
0
.
8955
−
0
.
4444
0
−
0
.
0911
⎣
⎦
=
⎣
⎦
⎣
⎦
=
⎣
⎦
⎣
⎦
−
c
2
s
3
c
3
0
0
.
4443
0
.
8958
0
−
0
.
1048
s
2
01
−
0
.
025
0
1
−
2
.
909
⎡
⎤
−
0
.
0350
−
⎣
⎦
=
.
1344
−
2
.
907
Similarly for frame 5,
⎡
⎤
⎡
⎤
ω
x
ω
y
ω
z
−
0
.
1646
⎣
⎦
=
⎣
⎦
−
0
.
0249
−
2
.
637
and for frame 7,
⎡
⎤
⎡
⎤
−
0
.
0542
−
ω
x
ω
y
ω
z
⎣
⎦
=
⎣
⎦
0
.
13621
−
3
.
206
We now calculate the angular accelerations of the segment where
t
is
the sampling period:
α
x
(
fr
.
6
)
=
[
ω
x
(
fr
.
7
)
−
ω
x
(
fr
.
5
)
]
/
2
t
=
[
−
0
.
0542
−
(
−
0
.
1646
)
]
/
0
.
03333
=
3
.
312 r
/
s
2
α
y
(
fr
.
6
)
=
[
ω
y
(
fr
.
7
)
−
ω
y
(
fr
.
5
)
]
/
2
t
=
[
−
0
.
1362
−
(
−
0
.
0249
)
]
/
0
.
03333
3
.
337 r
/
s
2
=−
α
z
(
fr
.
6
)
=
[
ω
z
(
fr
.
7
)
−
ω
z
(
fr
.
5
)
]
/
2
t
=
[
−
3
.
206
−
(
−
2
.
637
)
]
/
0
.
03333
17
.
07 r
/
s
2
=−
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