MECHANICAL WORK, ENERGY, AND POWER - Biomechanics and Motor Control of Human Movement - page 167

6.3.2 External Work Calculation
It was noted in Sections 6.3.1.4 and 6.3.1.5 that the work calculations done
using Equations (6.6) and (6.25) automatically take into account all work done
by the muscles independent of whether that work was internal or external.
There is no way to partition the external work except by taking measurements
at the interface between human and external load. A cyclist, for example,
would require a force transducer on both pedals plus a measure of the velocity
of the pedal. Similarly, to analyze a person lifting or lowering a load would
need a force transducer between the hands and the load, or an imaging record
of the load and the body (from which an inverse solution would calculate the
reaction forces and velocity). The external work W e is calculated as:

t 2

W e = F r · V c dt (6.26)

t 1

=

where F r reaction force vector, newtons

V c = velocity of contact point, m/s

t 1 , t 2 = times of beginning and end of each power phase

6.4 POWER BALANCES AT JOINTS AND WITHIN SEGMENTS

In Section 6.0.2, examples were presented to demonstrate the law of conser-
vation of mechanical energy within a segment. Also, in Section 6.0.6, muscle
mechanical power was introduced, and in Section 6.0.9, the concept of pas-
sive energy transfers across joints was noted. We can now look at one other
aspect of muscle energetics that is necessary before we can achieve a com-
plete power balance segment by segment: the fact that active muscles can
transfer energy from segment to segment in addition to their normal role of
generation and absorption of energy.

6.4.1 Energy Transfer via Muscles
Muscles can function to transfer energy from one segment to the other if the
two segments are rotating in the same direction. In Figure 6.18, we have two
segments rotating in the same direction but with different angular velocities.
The product of Mω 2 is positive (both M and ω 2 have the same polarity), and
this means that energy is flowing into segment 2 from the muscles responsible
for moment M . The reverse is true as far as segment 1 is concerned, Mω 1 is
negative, showing that energy is leaving that segment and entering the mus-
cle. If ω 1 = ω 2 (i.e., an isometric contraction), the same energy rate occurs

and a transfer of energy from segment 1 to segment 2 via the isometrically
acting muscles. If ω 1 > ω 2 , the muscles are lengthening, and thus, absorption