Arm control equations, with exercize

[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

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

  frf/rsr = g1/g2


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.

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

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.

Bill P.