[From Bill Powers (920307.0900)]

I glossed over the bug leg details a bit in the previous post.

Consider two opposing muscles operating a limb around one joint.

A neural signal shortens a muscle by making its contractile component
shorter. A force is developed if the spring compnent is stretched relative
to the muscle's shortened state. We can lay out two muscles in a straight
line, anchored at the ends (with Xs), to show how a balanced pair of
muscles responds to a balanced pair of signals. The center point (marked O)
moves toward the side with the larger signal.

large signal small signal
> >
X//////////O/ / / / / / /X
>
>
position of
center, no

If you imagine a pulley at the position of O, and the two springs draped
over it, you can convert to angular motion around a pivot. The distance x

If an external force is now applied at O, positive to the right, the center
will be deflected to the right by an amount depending on the passive spring
constant of the muscle, k:

small signal large signal
> >
X/ / / / / /O////////X
>
> Fr <-- (restoring force)
> Fa ---> (external force)
x0| x
> ----->|
> position of
> center with
> deflecting force
>
position of
center, no

If S is the difference between left and right signals, then x0 = cS, where
c is the contraction constant (in radians per nervous system unit or NSU).
S represents (right - left) signal here and positive distances and forces
are measured to the right. The zero-point of x0 is at the midpoint between
the anchors.

If k is the spring constant of the combined muscles in force per radian of
stretch, then restoring force is

Fr = -k(x - x0), or
Fr = -k(x - cS).

Applied force plus restoring force equals zero, so

Fa - k(x - cS) = 0.

The deflection is then a function of the Signal and the external applied
Force:

x = (Fa/k + cS)

This is the "massless" version, which will be reasonably realistic.

If the leg has mass (moment of inertia) and friction, we have a
differential equation relating applied force and deflection x:

Fr = k(x - cS) + f(dx/dt) + m(d2x/dt^2) or with an applied force Fa,

Fa - k(x - cS) - f(dx/dt) - m(d2x/dt^2) = 0

where f = coefficient of friction, and "m" is really the moment of inertia
about the joint. It actually gets a lot more complex than that, which is
why I don't recommend using a completely real physical model without a
pretty high-powered mechanicist helping out.

···

--------------------------------------------------------------
Best,

Bill P