ANTHROPOMETRY - Biomechanics and Motor Control of Human Movement - page 91

Example 4.3. (a) A prosthetic leg has a mass of 3 kg and a center of mass
of 20 cm from the knee joint. The radius of gyration is 14.1 cm. Calculate I
about the knee joint.

mρ 0 = 3 ( 0 . 141 ) 2 m 2 I 0 = = 0 . 06 kg ·

= I 0 + mx 2
=

I

3 ( 0 . 2 ) 2 m 2 0 . 06 + = 0 . 18 kg ·

(b) If the distance between the knee and hip joints is 42 cm, calculate I h
for this prothesis about the hip joint as the amputee swings through with a
locked knee.

x = distance from mass center to hip = 20 + 42 = 62 cm

mx 2 I = I 0 +

3 ( 0 . 62 ) 2 m 2 = 0 . 06 + = 1 . 21 kg ·

Note that I h is about 20 times that calculated about the center of mass.

4.1.7 Use of Anthropometric Tables and Kinematic Data
Using Table 4.1 in conjunction with kinematic data we can calculate many
variables needed for kinetic energy analyses (see Chapters 5 and 6).This table
gives the segment mass as a fraction of body mass and centers of mass as a
fraction of their lengths from either the proximal or the distal end. The radius
of gyration is also expressed as a fraction of the segment length about the
center of mass, the proximal end, and the distal end.

4.1.7.1 Calculation of Segment Masses and Centers of Mass

Example 4.4. Calculate the mass of the foot, shank, thigh, and HAT and its
location from the proximal or distal end, assuming that the body mass of the
subject is 80 kg. Using the mass fractions for each segment,

Mass of foot = 0 . 0145 × 80 = 1 . 16 kg

Mass of leg = 0 . 0465 × 80 = 3 . 72 kg

Mass of thigh = 0 . 10 × 80 = 8 . 0kg

Mass of HAT = 0 . 678 × 80 = 54 . 24 kg

Direct measures yielded the following segment lengths: foot = 0 . 195 m,

leg = 0 . 435 m, thigh = 0 . 410 m, HAT = 0 . 295 m.