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kwon3d
01-29-2004, 11:51 AM
Dear Jonas, Richard, and all,

I have a slightly different idea from that of Richard. A dot product of
F * V can be expressed in several different ways:

F * V = f.v.cosA [1]
F * V = ( f.cosA ).v [2]
F * V = f.( v.cosA ) [3]

where f = magnitude of force, v = magnitude of velocity, and A = angle
between the vectors. Eqs. 1-3 show the same thing with different
locations of cosA. Among the equations, Eq 3 basically says you only
consider the compont of V in the direction of F. F * V can be rewritten
as

F * V = F * (Vf + Vn)
= F * Vf + F * Vn
= F * Vf + 0 [4]

where Vf = velocity in the direction of the force, Vn = velocity normal
to the force. The second term in Eq 4 becomes 0 because cos 90 = 0. Eq 4
is identical to Eq 3. Similarly,

F * V = ( Fi + Fj + Fk ) * ( Vi + Vj + Vk )
= Fi * ( Vi + Vj + Vk ) + Fj * ( Vi + Vj + Vk ) + Fk * (
Vi + Vj + Vk )
= ( Fi * Vi + Fi * Vj + Fi * Vk ) + ( Fj * Vi + Fj * Vj
+ Fj * Vk ) + ( Fk * Vi + Fk * Vj + Fk * Vk )
= ( fi.vi + 0 + 0 ) + ( 0 + fj.vj + 0 ) + ( 0 + 0 +
fk.vk )
= fi.vi + fj.vj + fk.vk [5]

where i, j, k = orthogonal axes. Eq 5 is another equation of dot product
we can use when both vectors are DESCRIBED IN THE SAME SYSTEM and the
system is ORTHOGONAL. The dot product of two vectors can have 9 scalar
terms but 6 of them happens to be 0 because some force and velocity
components are not related (perpendicular to each other). The point is
that Eq 5 is simply a byproduct of Eqs 1-3.

Now, let's go back to the issue of muscle power (M * W). Let's say we
are studying the muscule power at the shoulder. We can define two
reference frames here: the trunk frame and the upperarm (UA) frame. If
both the muscle moment and the angular velocity of the joint are
described in the trunk frame, M * W can be written as:

M * W = M * ( Wi + Wj + Wk )
= M * Wi + M * Wj + M * Wk
= mi.wi + mj.wj + mk.wk [6]

where i, j, k = trunk frame axes (orthogonal). What Eq 7 basically shows
is how the muscle moment is associated with the three orthogonal angular
velocity components described in the trunk frame. This is also equal to

M * W = M * ( Wi' + Wj' + Wk' )
= M * Wi' + M * Wj' + M * Wk' [7]

where i', j', k' = the UA frame axes. Eq 7 shows how the muscle moment
is associated with the three orthogonal angular velocity vectors
described in the UA frame. The results from Eqs 6 and 7 must be the
same:

M * Wi + M * Wj + M * Wk = M * Wi' + M * Wj' + M * Wk'
[8]

Of course, this doesn't mean

M * Wi = M * Wi'
M * Wj = M * Wj'
M * Wk = M * Wk'

Eqs 6-7 show different ways to associate the muscle moment with the
angular velocity. In Eq 8, each dot product can generate 3 non-zero
scalar terms because the two vectors are not described in the same
reference frame.

Yet, there is another way to write M * W:

M * W = M * ( W1 + W2 + W3 )
= M * W1 + M * W2 + M * W3 [9]

where W1, W2, W3 = three Kardanian angular velocity vectors. These
vectors are not orthogonal. The three terms in Eq 9 shows how the muscle
moment is associated with the three Kardanian rotations: how the moment
is associated with the flexion/extension velocity of the shoulder, etc.
Each term in Eq 9 can generate up to 3 non-zero scalar terms. But still

M * Wi + M * Wj + M * Wk = M * W1 + M * W2 + M * W3 [10]

Again, this doesn't mean

M * Wi = M * W1 ...

W1, W2, and W3 can be described in either the trunk reference frame or
the UA reference frame. Visit the Orientation Angles vs. Angular
Velocity page (http://kwon3d.com/theory/euler/avel.html) for details.

Back to what Richard stated:

> = ai.bi + aj.bj + ak.bk +
(aj.bi + bj.ai).(i.j) +
(ak.bi + bk.ai).(i.k) +
(ak.bj + bk.aj).(j.k)

If one is obsessed with the products of the components (ai.bi + aj.bj +
ak.bk) described in different systems (orthogonal or nonorthogonal), the
last three lines in the above equation are problematic. However, if you
think about the essence of the dot product (how to associate the
vectors), each of the three lines in the following equation are what are
meaningful:

> (a.b) = ai.bi.(i.i) + ai.bj.(i.j) + ai.bk.(i.k) +
aj.bi.(j.i) + aj.bj.(j.j) + aj.bk.(k.k) +
ak.bi.(k.i) + ak.bj.(k.j) + ak.bk.(k.k)

In my opinion, the angular velocity vector does not have to be broken
down into three orthogonal vectors. As long as W = W1 + W2 + W3, M * W
does not change (invariant). The only question is whether breaking down
W into W1, W2, and W3 is biologically meaningful. W1, W2, and W3 are the
three independent Kardanian rotation vectors and, to me, it is perfectly
OK.

Thank you for reading another lengthy posting!

Young-Hoo
------------------------------------------------------
- Young-Hoo Kwon, Ph.D.
- Biomechanics Lab, Texas Woman's University
- kwon3d@kwon3d.com
- http://kwon3d.com
------------------------------------------------------

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