THREE-DIMENSIONAL KINEMATICS AND KINETICS - Biomechanics and Motor Control of Human Movement - page 193

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 =−