[From Bill Powers (931003.1615 MDT)]

Avery Andrews (931003.1030) --

Your derivation of apportionment of d.f. among joints checks out

with me. As an alternative you can use the quasi-static approach

and get the same result:

(1)phr = rfr + rsr (forearm + shoulder ref rotations,

perception = reference)

(2) ehr = rhr - phr (ref hand rotation - perceived hand rot)

(3) rfr = g1 * ehr

(4) rsr = g2 * ehr (ref forearm & shoulder rot driven by

error signal)

solve for perceived hand rotation phr:

phr = g1 * (rhr - phr) + g2 * (rhr - phr), or

phr = (g1 + g2)*rhr/(1 + g1 + g2);

lim (phr) = rhr.

g1 + g2 --> infinity

So the hand rotation matches the reference rotation for infinite

gain.

From (3) and (4) it follows trivially that

frf/rsr = g1/g2

## ···

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The unused degree of freedom comes into play if we assume a

higher-level control system which can add a reference signal to

rfr. Let the reference signal be rf. We modify equation (3) to

add the effect of this extraneous signal to the output of the

hand-rotation control system:

(5) rfr = g1 * ehr + rf

Now the solution for ehr is

ehr = (rhr - rf)/(1 + g1 + g2).

Substituting this into the equations for rsr and rfr we have

rsr = g1 * (rhr - rf)/(1 + g1 + g2) and

rfr = g2 * (rhr + rs)/(1 + g1 + g2)

If g1 = g2 = g >> 1, we have

rsr = (rhr - rf)/2 and

rfr = (rhr + rf)/2

So an arbitrary rotation of the forearm by the amount rf/2 is

compensated by an equal and opposite rotation at the shoulder,

keeping the hand angle the same.

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Here's a "yoga" exercise illustrating the independence of rsr and

rfr over some range. This exercise uses rotations about the

longitudinal axes of forearm and upper arm with the arm

completely extended. The same equations would apply.

(1) Hold your right arm out straight in front of you with the

palm up and horizontal. By rotating JUST your forearm, you can

now flip the hand by about 180 degrees counterclockwise. Grasp

your elbow joint with your left hand to feel that the upper arm

does not rotate (much). Your right hand is now palm down.

(2) Now rotate the right hand an additional 90 degrees

counterclockwise (or as far as it will go). You now find that the

upper arm must do the rotating to get this additional range. The

hinge at the elbow joint has now rotated counterclockwise by

around 90 degrees.

(3) Finally, rotate the hand 90 degrees clockwise (to the palm-

down position), without changing the rotation of the elbow joint

hinge. The hand (and forearm) is now in the same position as at

the end of step (1), but the upper arm is rotated 90 degrees

counterclockwise from its position at the end of step (1).

(4) You should now be able to rotate the upper arm back to its

original position while holding the hand palm down. This may take

considerable practice. Even more practice may be needed to start

with the elbow hinge in the position at the end of step (1), and

rotate the upper arm 90 counterclockwise while keeping the hand

in the palm-down horizontal position.

(5). If you start with the hand palm-up and rotate it smoothly

through 270 degrees counterclockwise (position at end of step

three), you may see that this motion is apportioned between upper

arm and forearm so the two movements take the same time to go

through their ranges. The forearm rotates counterclockwise by 180

degrees while the upper arm rotates counterclockwise by 90

degrees.

The angular measures are approximate, and there may be some

interactions that you can't prevent.

All this says that the control systems involved are multiple and

can be used in various combinations depending on what you're

trying to do. Our simple models obviously leave out many

complications -- and capabilities.

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Best,

Bill P.