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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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

------------------------------------------------------

-----------------------------------------------------------------

To unsubscribe send SIGNOFF BIOMCH-L to LISTSERV@nic.surfnet.nl

For information and archives: http://isb.ri.ccf.org/biomch-l

Please consider posting your message to the Biomch-L Web-based

Discussion Forum: http://movement-analysis.com/biomch_l

-----------------------------------------------------------------