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Â Bruce Abbott (2015.01.28.2210 EST)

RM: When you talk about the counter force having a feedback effect I assume that you mean that the counter force has an effect on the cause of that force. The cause of that force is the deviation, d, of the mass from it’s resting position, r, per:Â

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F = k*(r-d) Â Â Â Â Â Â Â Â Â (1)

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Â Â BA: The deviation, d, should equal r-p, where p is the current position.

RM: I meant d to represent what you are calling p. Equation 1 is just Hooke’s law: restoring force, F, is proportional, by the spring constant, k, to the deviation (r-d) of the position of the mass, d, from it’s resting position, r. Â

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RM: where F is the counter force and s the spring constant. The counter force has a feedback effect if that force has an effect on its cause, which is r-p.

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BA: r-p is the deviation.Â The cause is some force that is acting to compress the spring.Â Iâ€™ve referred to this force as a disturbance.Â Since when does the output of a negative feedback system feed back to affect the disturbance?

RM: Since never. In my notation r-d (which is written as r-p is the piece you quote – another mistake of mine – Â is the deviation of the mass from it’s resting position, same as your r-p. There is negative feedback if the restoring force, F, affects the deviation, r-d for me ( r-p for you) Â that caused it.Â

RM: Taking resting position, r, as the 0 point of mass position, the counter force would have a feedback effect if:

d = -g*F + Fp Â Â Â Â Â Â (2)Â

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BA: Now you are defining d as the difference between two forces.Â Above it was the difference between two positions.

RM: No, d is still the position of the mass, measured as a deviation from the the resting state. I see now that this is lousy notation on my part so let me redefine the deviation of the mass on a spring from its resting position as the variable x, which is r-p (using your notation). Then the “forward” equation for the mass-spring system is, again, Hooke’s law, now written as

F = k*x Â Â Â Â Â Â Â Â Â Â (1)Â

and now equation 2 above is correct as the feedback equation, with x now replacing d:

x = -g*F + Fp Â Â Â Â Â Â (2)Â

What equation 2 says, then, is that a deviation, x, of the mass from its resting state is caused by an external force (Fp) – your pull up or down on the mass – and the feedback effect of the restoring force, F, if such a feedback connection exists (if g<>0).Â

RM: What I am saying is that in the mass-spring system g is actually 0 so there is **no feedback** from F to x (the cause of F); so the mass-spring system is an open loop system, like the pendulum, the marble rolling to the bottom of a bowl, the raindrop running to the edge of a tilted window and William James’ iron filings running to the magnet.Â

RM: So to determine whether the counter force, F, has a feedback effect we have to know how much Fp (the force applied to the mass) would be expected to displace the mass if there were no counter force feedback. This can be derived using equation (1) and solving for d as a function of Fp (instead of F). Letting r again be 0 we get

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d = -1/k Fp Â Â (3)

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RM: So if the counter force, F, has a feedback effect, applying force Fp to the mass will produce far less deviation from the resting state than expected. This is indeed what you would find if there were a control system actively varying F to compensate for the disturbance, Fp, to d. But I think that what you would find if you pulled down, say, with force Fp on a mass attached to a spring of known spring constant, k, is that the resulting deviation of the mass from it’s resting position is exactly what is expected from equation 3.Â

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BA:Â Well, garbage in, garbage out.

RM: I think it becomes less like garbage if you substitute my new notation, x, for d in equation 3. All I’m saying above is that **if** there is negative feedback in a mass-spring system there will be disturbance resistance. So this is a way of testing whether or not there is, indeed, disturbance resistance in a mass-spring system. It’s the test for the controlled variable.Â

RM: Equation 3 gives the *expected* deviation of the mass from its resting state when the disturbance force Fp is applied to the mass and there is **no feedback**!Â If when the disturbance force is applied the *observed* deviation (call it x’) is equal to the expected deviation (x) that is evidence that there is no negative feedback in the system.Â

Â RM: I’m sure this produces a nice harmonic motion of the position variable. But it’s hard for me to see anything that looks like a feedback loop. Perhaps what you see as a feedback loop is effect of the position and velocity of the mass on itself via the spring constant and damping constants. But I see these as simply delay loops, showing how the position and velocity of the mass at time t should be turned into a position and velocity of the mass at time t + dt.Â

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BA:Â Now this is getting downright bizarre.Â You would have us believe that a system diagram with rate and position negative feedback (opposing the acceleration-of-mass vector) is not a diagram of a negative feedback system.

RM: The diagram you posted is of a system that produces a damped oscillation of a mass on a spring. The system is doing with digital simulation what can be done with differential equations. In both cases you can simulate the behavior or a mass on a spring that returns to its original position (equilibrium point) after a deflection from that point using the open - loop equations for restoring force, per equation 1, and damping, c(dx/dt), where c is the damping constant.Â

RM: As I understand it your digital simulation produces the dynamic behavior that would be produced by the differential equation that defines a damped oscillation of a mass spring system:Â

Â Â Â Â Â Â Â Â Â m(d2x/dt2) + c(dt/dt) + kx = 0 Â Â Â Â Â Â Â Â Â (4)Â

without having to solve the differential equation analytically. I don’t believe that there is any negative feedback involved in equation 4 but I’m not much on differential calculus. But I know that **Richard Kennaway** is quite good at it and he should be able to tell us whether there is any feedback (positive or negative) involved in the differential equation (4) that defines the behavior of a damped oscillation of a mass on a spring that occurs after it is displaced by an amount x.

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RM: What you have here is a lineal causal system that operates over time. There is no feedback in the control theory sense since the force generated by the changes in position and velocity have no *simultaneous* Â effect on the the causes of that force.Â

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BA: You are asserting that compression of a spring by a force does not simultaneously produce a counterforce as the spring pushes back.Â It does in my world . . .

RM: No, I’m asserting that this counter force is **not feedback**; it is simply a force caused in response to the displacement, x, of the mass. F = kx. That force is not “countering” or “restoring” anything until the disturbance (Fp) that caused the displacement, x, is removed. When the disturbance is removed the “counter force”, which was **potential energy** when the displacement occurred, becomes **kinetic energy** that combines with frictional (damping) forces to move the mass back (dynamically, according to equation 4) to its pre-displacement position **after the disturbance is removed**.Â The disturbance was completely effective when applied; the system returns to it’s resting point when the disturbance is removed, just as is predicted from the lineal, open loop causal model of physics (equation 4).Â

RM: Another way to see that F is not really a countering force is to note that the displacement of the mass from its resting state will remain equal to x as long as the disturbing force, Fp, is being applied. That is, the “counter force” does no “countering” while the disturbance is acting; the “countering” doesn’t occur until the disturbance is removed. So there is really no countering.

BA:Â Â I donâ€™t know why you added â€œthere is no controlâ€? to the end of your last sentence.Â It suggests that I am asserting that passive equilibrium systems are control systems, which I have taken pains to make clear they are not.Â (They do, however, reduce the effects of disturbances to some degree.)

RM: Ignoring for the nonce the fact that the mass-spring system does not reduce the effects of disturbances at all, how in the world do you distinguish “reducing the effects of disturbance” from " control". **Control is defined as reducing the effects of disturbances**. AÂ good control system reduces the effects of disturbances almost completely; a poor control system reduces the effects of disturbances "to some degree’.Â

RM: So if these equilibrium systems really did “reduce the effects of disturbances to some degree” then they would be control systems; crumby control systems, but control systems nonetheless. So I said that there is **no control** done by equilibrium systems like the mass spring system because, in fact, Â they don’t reduce the effects of disturbance at all. When you apply a force disturbance, Fp, to a mass on a spring with spring constant k that mass is displaced by exactly the amount expected on physical grounds (x = 1/kFp). There is absolutely no reduction at all in the effect of that disturbance on the purported controlled (or stabilized, if you prefer) variable.Â

BA:Â Well, hereâ€™s my last try at convincing you.Â I discovered one last arrow in my quiver.Â Here it is:

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procedure TMainForm.RunModel;

begin

Â // acceleration = applied force minus velocity and spring feedback, over mass

Â Accel := (Force + ForceDist - D*Vel - S*Pos)/M;

Â LastPos := Pos;Â Â Â // save position for computing velocity

Â // New position = old position + velocity*dt + 1/2 acceleration*dt-squared

Â Pos := Pos + (Vel + 0.5*Accel*dt)*dt;

Â Vel := Vel + Accel*dt;Â Â // update velocity; dt is time step per iteration

Â VelFB := (Pos - LastPos)/dt;Â // compute velocity from delta-pos

Â e3 := SepRef - (Pos - Target);Â // pos - target is actual separation

Â e2 := PosGain * e3 - velfb;Â // posgain*e3 is output of position control system

Â Force := VelGain * e2;Â Â Â Â Â Â // which is the velocity reference signal

end;

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This is a control system acting on a mass-spring-damper system.Â Â If you set PosGain to zero and VelGain to zero, then you have the passive equilibrium system, where ForceDist is the applied disturbance acting on the mass.Â Notice the label over the Accel variable:

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// acceleration = applied force minus velocity and spring feedback, over mass

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This code snippet is from the NonAdaptive program described in LCS III.Â The comments are by Bill Powers.Â Velocity and spring feedback, subtracting from applied force.Â That makes it *negative* feedback.

RM: The “spring feedback” of which Bill speaks is the variable Force. It is an actual feedback variable inasmuch is it has an effect on the variable that caused it, the variable Pos, and, ultimately, on itself. So Force has an effect on Accel which has an effect on Pos which has an effect on Force. There is a control loop and Force is part of it.Â

RM: So Force is appropriately called the “spring feedback” because it is the force exerted by the spring on the the cause of that force – the deviation of the position of the mass from the position reference. If you set PosGain and/or VelGain to 0 you remove the “spring feedback” (Force) and there is no more control system involved.

RM: I don’t see any evidence of the computation of a restoring force in this code. Perhaps the restoring force is ForceDist, which is apparently computed outside the procedure and may be computed as a function of the change in position of the ends of the muscle produced by the tendons. Is that it?Â

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BA: Time to concede, Rick.

RM: I would happily concede if I saw any evidence at all that I was wrong. But this discussion is really helping me understand what’s going on with this equilibrium stuff. I think equilibrium models are simply an attempt to use a causal model (Newton’s laws of motion) to account for purposeful (control) behavior. But I’d love you to show me that I’m wrong. You’ve done it before; maybe you can do it again!Â

BestÂ

Rick