Page 231 - Biomedical Engineering and Design Handbook Volume 1, Fundamentals

208 BIOMECHANICS OF THE HUMAN BODY

New position vectors designating the locations of point G and D are given as
2
1
r = l 4 + r C (8.58)
2
G 2
and r = r + l 4 (8.59)
C
D
respectively for the new segment.
When solving a three-segment system as the one seen in Fig. 8.5, the kinetic and potential energy
must be determined independently for each segment. Since the kinetic and potential energy was
solved previously for segment BC and is given in Eqs. (8.48) and (8.49), consideration needs to be
given only to segment CD.
In a similar manner as before, the angular velocity vector of segment CD is determined to be

ω =− Ωcos θ c + Ωsin θ c + θ c (8.60)
21
2 2
C
2 3
where θ is the angle between the segment and the vertical and θ 2 is the time rate of change of the
2
angle. Also, the components for the mass moment of inertia about point G in the c , c , c frame of
1
3
2
2
reference are
I = 0
c 1 (8.61)
1 2
and I = I = m l (8.62)
24
c 2
12
c 3
and the kinetic energy of segment CD is defined by the equation
1
T = ( I ω 2 + I ω 2 + I ω 3 ) + 1 m v • v (8.63)
2
2 c 1 c 1 c 2 c 2 c 3 c 3 2 2 G 2 G 2
where the velocity vector at point G is
2
v = v + ω × r (8.64)
/
G 2 C c G C
2
In order to solve Eq. (8.64), the velocity vector at point C and the cross product must be deter-
mined.
v = v + ω b × r CB /
B
C
so that
b + θ
v =− Ω b + − Ωcosθ 1 1 1 2 1 3 ) l 3 1 (8.65)
b + Ωsinθ
l
b )
b × (
(
2
3
C
v v = θ b − Ω( l + l sin θ 1 b ) 3
l
C
3
31 2
2

and ω × r G C = − ( Ω cos θ c + Ω sin θ c + θ c ) × ( l 2 4 c ) (8.66)
2 2
1
C
2 3
21
/
2
l 4 l
ω ω × r G C = 2 θ c − 2 4 Ω sin θ c
/
C
23
22
2
so that
l
v = l θ b − Ω( sin θ + l ) b + l 4 θ c − l 4 Ωsinnθ c (8.67)
G 2 31 2 3 1 2 3 22 23
2 2
Note that Eq. (8.67) contains velocity terms from both moving coordinate systems b , b , b and c ,
1 2 3 1
c , c and should be represented entirely in terms of c , c , c Therefore, a coordinate transformation
2 3 1 2 3.
matrix is defined that will manage this.