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vzatsiorsky65
01-29-2004, 02:56 AM
Dear Richard and all:

I agree with what you wrote and want to add a simple 'geometric' explanation
of what is happening when power is computed in the ICS axes. To make the
understanding easier I replaced your vector a with F (force) and vector b
with V(velocity).

In the JCS system, the axes i and k are in the plane that is orthogonal to
axis j (a floating axis). Let's introduce in this plane an axis l that is
orthogonal to axis i and consider an orthogonal system of coordinates ijl.
In this reference system, power FV can be represented as a sum of three
terms associated with the individual force-velocity projections on the
coordinate axes:

FV=fi*vi+fj*vj+fl*vl

Because power is invariant [its magnitude does not depend on the selected
(inertial) system of coordinates] and the first two terms in the above
equation are equally valid for the JCS system, the following equality is
also valid:
fl*vl= fk*vk+(fk*vi+fi*vk) (i.k)

In this equation: fl*vl is a power term representing the power associated
with the force and velocity components along the orthogonal axis l; fk*vk
is a similar term representing the power associated with the non-orthogonal
axis k, (fk*vi+fi*vk) (i.k) is the difference (DELTA) between the above
power values. Because (i.k.) is simply a cosine of angle A formed by the
axes i and k, the DELTA = 0 when k is along axis l (i.e. the system ijk is
orthogonal) and DELTA is not equal to zero in all other cases. DELTA=1 when
axes i and k are along the same direction. In the latter case the
singularity occurs.

Sincerely,

----- Original Message -----
From: "Richard Baker"
To:
Sent: Wednesday, January 28, 2004 5:49 PM
Subject: Re: [BIOMCH-L] More angles and powers

> Dear Jonas and all,
>
> Ton and I have had a brief exchange of notes on a sub-topic of the current
> discussion - that of whether the decomposition of the dot product a.b
> =ax.bx+ay.by+az.bz works in non-orthogonal axis system. I resorted to an
> old textbook on mathematical physics which said "no" with some fairly
> math to explain. Ton's come up with the same answer but in a much more
> insightful manner:
>
> Let a and b be vectors, and i,j,k be unit vectors along the coordinate
axes
> which may be non-orthogonal. Let ai, aj, ak be the scalar components
along
> each axis. Then:
>
> a = ai.i + aj.j + ak.k
> b = bi.i + bj.j + bk.k
>
> Hence:
>
> (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)
>
> = ai.bi + aj.bj + ak.bk +
> (aj.bi + bj.ai).(i.j) +
> (ak.bi + bk.ai).(i.k) +
> (ak.bj + bk.aj).(j.k)
>
> In the JCS, axes 1 and 2 are orthogonal, and axis 2 and 3 are orthogonal,
> so we lose the cross terms with (i.j) and (j.k). The (i.k) term
> remains :-(. So it seems that you need to add this term if you wanted
> to compute total power from JCS angular velocities and JCS moment
> components.
>
> We are both agreed that vector relationships must hold whatever the
> co-ordinate system used (whether orthogonal or not) but the way in which
> these are calculated from the basic components will depend on the
> characteristics of the co-ordinate system.
>
> Richard
>
>
> Richard Baker
>
> Gait Analysis Service Manager, Royal Children's Hospital
> Flemington Road, Parkville, Victoria 3052
> Tel: +613 9345 5354, Fax +613 9345 5447
>
> Adjunct Associate Professor, Physiotherapy, La Trobe University
> Honorary Senior Fellow, Mecahnical and Manufacturing Engineering,
Melbourne
> University
>
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