Homeostasis

[From Bruce Abbott (2015.01.27.1010 EST)]

Rick Marken (2014.01.26.1600) –

Martin Taylor (2014.01.23.13.35)

RM: I guess I don’t really understand the difference between control and homeostasis.

MT: As I understand it, the essence of a control loop is the asymmetry between the input side and the output side. Homeostasis can function with any kind of negative feedback loop…

BA: If homeostasis includes any stable negative feedback system, then the term covers both control systems and passive equilibrium systems. The two kinds of systems differ in loop gain.

BA: In control systems, the loop gain is greater than 1, whereas in the passive equilibrium system it is greater than zero but less than or equal to 1. Put another way, the control system draws on its own external source of energy to counteract the effects of disturbances (which is why the gain can be greater than 1), whereas the passive equilibrium system uses the energy supplied by the disturbance itself (which is why the gain cannot be greater than 1).

BA: An example of the latter is the spring. As a spring is compressed by an external force (disturbance), that compression generates a counterforce in the spring. The spring continues to compress until the counterforce equals the force (a new equilibrium). When the force is withdrawn, the energy stored in the spring returns the spring to its former length (its former equilibrium value). In the body, chemical reactions in the liver act to remove excess glucose (blood sugar) from the bloodstream by converting it to glycogen (a starch) and storing it in the liver. As glucose levels in the bloodstream fall, the liver converts the glycogen back to glucose and releases it into the blood. I don’t know whether these reactions require an external source of energy or simply depend on the concentration of glucose, but if the latter is true, then this homeostatic system is a passive equilibrium system.

Bruce

[From Rick Marken (2015.01.27.1100)]

···

Bruce Abbott (2015.01.27.1010 EST)–

BA: If homeostasis includes any stable negative feedback system, then the term covers both control systems and passive equilibrium systems. The two kinds of systems differ in loop gain.

RM: I don’t think equilibrium systems are negative feedback systems. There is no “feedback”; no loop.

BA: In control systems, the loop gain is greater than 1, whereas in the passive equilibrium system it is greater than zero but less than or equal to 1. Put another way, the control system draws on its own external source of energy to counteract the effects of disturbances (which is why the gain can be greater than 1), whereas the passive equilibrium system uses the energy supplied by the disturbance itself (which is why the gain cannot be greater than 1).

BA: An example of the latter is the spring.

RM: I think the loop gain of a spring is 0; a spring is just a causal system. There is no feedback.

BA: As a spring is compressed by an external force (disturbance), that compression generates a counterforce in the spring.

RM: Isn’t that just the third law of motion?

BA: The spring continues to compress until the counterforce equals the force (a new equilibrium). When the force is withdrawn, the energy stored in the spring returns the spring to its former length (its former equilibrium value).

RM: RIght, it’s all just cause-effect. There is no loop at all. This is all handled rather well by good old Newtonian physics, isn’t it? Robert Hooke and all that.

BA: In the body, chemical reactions in the liver act to remove excess glucose (blood sugar) from the bloodstream by converting it to glycogen (a starch) and storing it in the liver. As glucose levels in the bloodstream fall, the liver converts the glycogen back to glucose and releases it into the blood. I don’t know whether these reactions require an external source of energy or simply depend on the concentration of glucose, but if the latter is true, then this homeostatic system is a passive equilibrium system.

RM: If glucose level were not a controlled variable we would be in big trouble, I believe. If glucose level were a variable in an “equilibrium system” like the length of a spring then glucose level would be “stabilized” only when the variables affecting glucose level behaved properly. For example, glucose level increases when we consume food just as the length of the spring increases when we pull on it. If glucose level were similar to spring length then it would return to its “correct” equilibrium state (the one that allows us to survive) only when we stopped eating for just the right amount of time. If we kept eating then glucose level would stay high just as the length of the spring would stay long if we kept pulling on it.

RM: Equilibrium systems are not not control systems; they are simply causal systems that can appear to do something like control if “control” is understood to be the return to an equilibrium state after removal of a “disturbance” and/or remaining in an equilibrium state when the “disturbance” is constant. That is, if “stability” is taken to be a form of control.

RM: Given your skill at building robots, Bruce, I think it would be great if you could build a robot that maintains, say, the angle of its arm using equilibrium or control. You could design it so that a user could switch from one to the other. Both would have variable “references” – in equilibrium mode variations in the reference would vary the “equilibrium point” and in control mode variations in the reference would vary the specification for perceived “equilibrium point”. So you couldn’t tell which mode the system was in by just looking at it’s behavior (variations in arm angle); you would have to apply disturbances to test to see whether you were dealing with an equilibrium or a control system.

Best

Rick


Richard S. Marken, Ph.D.
Author of Doing Research on Purpose.
Now available from Amazon or Barnes & Noble

[From Bruce Abbott (2015.01.27.2020 EST)]

Rick Marken (2015.01.27.1100)–

Bruce Abbott (2015.01.27.1010 EST)

BA: If homeostasis includes any stable negative feedback system, then the term covers both control systems and passive equilibrium systems. The two kinds of systems differ in loop gain.

RM: I don’t think equilibrium systems are negative feedback systems. There is no “feedback”; no loop.

BA: In control systems, the loop gain is greater than 1, whereas in the passive equilibrium system it is greater than zero but less than or equal to 1. Put another way, the control system draws on its own external source of energy to counteract the effects of disturbances (which is why the gain can be greater than 1), whereas the passive equilibrium system uses the energy supplied by the disturbance itself (which is why the gain cannot be greater than 1).

BA: An example of the latter is the spring.

RM: I think the loop gain of a spring is 0; a spring is just a causal system. There is no feedback.

BA: As a spring is compressed by an external force (disturbance), that compression generates a counterforce in the spring.

RM: Isn’t that just the third law of motion?

BA: The spring continues to compress until the counterforce equals the force (a new equilibrium). When the force is withdrawn, the energy stored in the spring returns the spring to its former length (its former equilibrium value).

RM: RIght, it’s all just cause-effect. There is no loop at all. This is all handled rather well by good old Newtonian physics, isn’t it? Robert Hooke and all that.

BA: Assume a mass-less spring supporting a mass M. Force (disturbance) acts on one end of spring. (The other end is fixed in position.) The applied force accelerates the end of the spring according to A = F/M. Acceleration changes the velocity from zero to some positive value in one time increment dt. This velocity changes the position of the spring-end but also generates a counterforce equal to the velocity times a damping constant. The change in position from the spring’s resting point generates a counterforce equal to k times the change in position, where k is the spring constant. The counterforce due to position and velocity combine to feed back on the mass in a direction that opposes the impressed force (i.e., negative feedback), reducing the acceleration. Iterate this loop until the system reaches a new equilibrium position.

If you don’t believe this, try implementing it in a spreadsheet. Don’t forget to include a leaky integrator somewhere in the loop to prevent the loop from blowing up the numerical integration.

The above is not a control system, but a control system is an equilibrium system. The difference lies in whether the energy to oppose the disturbance comes from the disturbance itself or from some independent source. The former reaches a passive equilibrium, the latter an active one.

Bruce

[From Rick Marken (2015.01.27.1940)]

···

Bruce Abbott (2015.01.27.2020 EST)

BA: Assume a mass-less spring supporting a mass M. Force (disturbance) acts on one end of spring. (The other end is fixed in position.) The applied force accelerates the end of the spring according to A = F/M. Acceleration changes the velocity from zero to some positive value in one time increment dt. This velocity changes the position of the spring-end but also generates a counterforce equal to the velocity times a damping constant.

RM: Isn’t that Newton’s third law of motion?

BA: The change in position from the spring’s resting point generates a counterforce equal to k times the change in position, where k is the spring constant. The counterforce due to position and velocity combine to feed back on the mass in a direction that opposes the impressed force (i.e., negative feedback), reducing the acceleration.

RM: Well, I’m no physicist but I think it’s the coils of the spring that make this situation different than just pushing on a mass. I think it’s easier to see what’s going on if we look at a pendulum rather than a spring. If you pull a pendulum bob away from its resting point and release it the bob will eventually return to its resting point just as the displaced mass on a spring will eventually return to its resting point. You could say that the change in the pendulum bob’s position generates a counterforce that opposes the change in position but it seems to non-physicist me that the counterforce is not really being “generated”; you have just changed the position of the bob relative to a force that’s already acting on the bob: gravity. So your displacement changes the direction of the force vector. So there is no negative feedback in the equilibrium behavior of a pendulum; the restoring force (gravity) is always the same, regardless of the amount of displacement of the bob. What changes is the orientation of the bob relative to gravity. As you will see if you read on, my intuitions turn out to be correct.

RM: I suspect that in the case of the mass-spring system the equivalent of gravity is the spring constant, k. Changing the position of the mass doesn’t “generate a counterforce” any more than displacement of the pendulum bob generates a counterforce. I think the restoring force that changes as a function of displacement of the mass is a property of the elasticity of the spring in the same way that the restoring force that changes as a function of displacement of the bob is a function of the orientation of the bob relative to gravity.

RM: Thanks to the internet I found that my intuition about the restoring force (what you call the counterforce) in the pendulum is correct. Moreover, it turns out that the cause of the restoring force in a displaced mass on a spring is exactly equivalent to that for a displaced pendulum.

RM: As you say above, the restoring force, F, for a displaced mass on a spring is proportional, by the spring constant, k, to the amount of displacement, delta x.

F = k* (delta x)

The equivalent equation for the restoring force of a displaced pendulum is:

F = - g * sine(theta)

where F is again the restoring force, g is gravity and theta is the angle of displacement of the bob from plumb.

RM: So you see in both cases the restoring force, F, is not really “generated” by displacement. In the case of the spring the force is generated by properties of the spring itself (the spring’s elasticity as measured by the spring constant k) and in the case of the the pendulum the force is generated by the mass of the earth (the gravitational constant g).

BA: The above is not a control system

RM: Nor is it a negative feedback system. There is no feedback in these systems. Feedback means that the output affects the input while, at the same time, the input affects the output. The equations for the restoring forces above show that, in these “equilibrium systems”, the output of the system (F) is affected by the input (displacement: delta x or theta). But the input is not, at the same time, affected by the output; the amount of displacement ( delta x or theta) is not in any way affected by the restoring force. That is, there is no relationship of the sort:

delta x = g (F)

or

sine (theta) = g (F)

These equilibrium systems are what Bill called Type Z (Zero Loop Gain) systems in his “Quantitative Analysis of Purposive Systems” paper (LCS I, p. 129). They are certainly interesting systems if you are interested in the behavior of non-living (non-purposive) systems but they don’t contribute anything but confusion to the study of living (purposive) systems.

, but a control system is an equilibrium system.

RM: Not really. A control system controls; an equilibrium system doesn’t control. A control system is organized as a closed negative feedback loop; an equilibrium system is organized as an lineal causal (Open Loop or Zero Loop Gain) system.

BA: The difference lies in whether the energy to oppose the disturbance comes from the disturbance itself or from some independent source.

RM: As you can see from the equations for restoring forces above, the energy to oppose what you call the disturbance (the displacement) does not come from the disturbance (displacement) itself; it comes from the spring elasticity (k) in the case of the mass on a spring and from gravity(g) in the case of the pendulum.

BA: The former [equilibrium system] reaches a passive equilibrium, the latter [control system] an active one.

RM: I’ll buy that.

Best

Rick


Richard S. Marken, Ph.D.
Author of Doing Research on Purpose.
Now available from Amazon or Barnes & Noble

[From Bruce Abbott (2014.01.28.0955 EST)]

Rick Marken (2015.01.27.1940)–

Bruce Abbott (2015.01.27.2020 EST)

BA: Assume a mass-less spring supporting a mass M. Force (disturbance) acts on one end of spring. (The other end is fixed in position.) The applied force accelerates the end of the spring according to A = F/M. Acceleration changes the velocity from zero to some positive value in one time increment dt. This velocity changes the position of the spring-end but also generates a counterforce equal to the velocity times a damping constant.

RM: Isn’t that Newton’s third law of motion?

BA: The change in position from the spring’s resting point generates a counterforce equal to k times the change in position, where k is the spring constant. The counterforce due to position and velocity combine to feed back on the mass in a direction that opposes the impressed force (i.e., negative feedback), reducing the acceleration.

RM: Well, I’m no physicist but I think it’s the coils of the spring that make this situation different than just pushing on a mass. I think it’s easier to see what’s going on if we look at a pendulum rather than a spring. If you pull a pendulum bob away from its resting point and release it the bob will eventually return to its resting point just as the displaced mass on a spring will eventually return to its resting point. You could say that the change in the pendulum bob’s position generates a counterforce that opposes the change in position but it seems to non-physicist me that the counterforce is not really being “generated”; you have just changed the position of the bob relative to a force that’s already acting on the bob: gravity. So your displacement changes the direction of the force vector. So there is no negative feedback in the equilibrium behavior of a pendulum; the restoring force (gravity) is always the same, regardless of the amount of displacement of the bob. What changes is the orientation of the bob relative to gravity. As you will see if you read on, my intuitions turn out to be correct.

BA: Wrong. In the case of the spring, compressing the spring does mechanical work, stretching bonds within the material of the spring and generating the counterforce pushing back against the compressive force. In the case of the pendulum, raising the bob by swinging the pendulum at its pivot increases the potential energy of the bob (obtained from work done by the disturbance to raise the bob) and changes the angle from which gravity is pulling on the bob relative to the pendulum shaft. In both cases the result is a counterforce that acts against the disturbing force and increases with the amount of displacement.

RM: I suspect that in the case of the mass-spring system the equivalent of gravity is the spring constant, k. Changing the position of the mass doesn’t “generate a counterforce” any more than displacement of the pendulum bob generates a counterforce. I think the restoring force that changes as a function of displacement of the mass is a property of the elasticity of the spring in the same way that the restoring force that changes as a function of displacement of the bob is a function of the orientation of the bob relative to gravity.

RM: Thanks to the internet I found that my intuition about the restoring force (what you call the counterforce) in the pendulum is correct. Moreover, it turns out that the cause of the restoring force in a displaced mass on a spring is exactly equivalent to that for a displaced pendulum.

. . . . .

RM: As you can see from the equations for restoring forces above, the energy to oppose what you call the disturbance (the displacement) does not come from the disturbance (displacement) itself; it comes from the spring elasticity (k) in the case of the mass on a spring and from gravity(g) in the case of the pendulum.

BA: Nope. Spring elasticity and the movement of the bob against the force of gravity are the mechanisms through which the force of the disturbance is converted to a counteracting restorative force. The force applied to compress the spring or raise the bob is the disturbing force.

BA: Here’s the bottom line: You can model these systems using the appropriate formulas from physics (as you did) or explicitly as negative feedback systems (as I did for the spring example).  You get the same system behavior either way. They are mathematically equivalent. Either way, you have a system in which a disturbing force generates an opposing counterforce in such as way as to produce the observed behavior: a spring that only compresses so far and springs back to its original length when the disturbance is removed, or a pendulum whose bob rises to a given position and then oscillates around its initial, straight down position when the disturbance is removed (pendulum released).

BA: The advantage of modeling these systems in terms of negative feedback is that you can then directly compare passive equilibrium systems to control systems. The spring-mass system, for example, can be modeled as having a reference value equal to the undisturbed position of the mass and a  loop gain that is less than or equal to one but greater than zero. This approach makes explicit the differences between such a passive equilibrium system and an actual control system. The passive system’s “reference� value is fixed and equal to the undisturbed value; the control system’s reference can be adjusted. The passive system’s loop gain cannot exceed 1.0 because its restorative energy comes from the disturbance itself; the control system’s loop gain can be extremely high because the energy it uses to generate the restorative force comes from an independent source. The passive equilibrium system will passively adjust to a continuous disturbance by moving to a new equilibrium value; the control system will generate whatever force is necessary (up to a limit) to counteract the disturbance and keep its controlled variable at or near the reference value.

Bruce

[From Rick Marken (2015.01.28.1120)]

···

Bruce Abbott (2014.01.28.0955 EST)Â

Â

BA: Wrong. In the case of the spring, compressing the spring does mechanical work, stretching bonds within the material of the spring and generating the counterforce pushing back against the compressive force. In the case of the pendulum, raising the bob by swinging the pendulum at its pivot increases the potential energy of the bob (obtained from work done by the disturbance to raise the bob) and changes the angle from which gravity is pulling on the bob relative to the pendulum shaft. In both cases the result is a counterforce that acts against the disturbing force and increases with the amount of displacement.

RM: OK, I can buy that. But if you consider the displacement a disturbance then the counterforce that is generated is not very effective since the disturbance is fully effective. I think the important concept here is that the counterforce generated by the displacement is potential energy; it’s not energy that is actively working against the disturbance, as it would be if this were actually a feedback system. So even if it’s correct to say that the displacement (calling it a disturbance is kind of a misnomer since the position of the mass is not a controlled variable) generates the counter force, that counter force is not acting against the displacement as the displacement occurs, as it would in a closed loop control system. Equilibrium systems are not closed loop systems!

BA: Here’s the bottom line: You can model these systems using the appropriate formulas from physics (as you did) or explicitly as negative feedback systems (as I did for the spring example).Â

RM: I don’t believe you did model the mass spring system as a negative feedback system; if you had then you would have had to model it with two simultaneous equations:

F = k*(r-d) Â Â (1)

d = -g*F + Fp    (2)

where F is the counter force, k is the spring constant, r is the resting position of the spring (assume it’s 0) , d is the actual position of the spring (when d<>0 there is displacement of the mass), g is the feedback constant (which I’ve made negative because you say this is a negative feedback system) and Fp is the force used to displace the mass (the equivalent of the disturbance variable, d, in PCT equations).Â

RM: Equation 1 is the “system” equation, relating input (position of the mass) to output (counter force generated by change in position). Equation 2 is the “environment” equation relating output (the counter force generated by your displacement of the mass) and environmental disturbances (your pull on the mass, Fp) to input.

RM: I think you will find that, in order to properly model the behavior of the mass on a spring, you would have to set the feedback constant, g, to 0, making the mass spring system an example of Powers’ Z system.That is, this “equilibrium system” is just a lineal causal system; there is no control involved at all. If there were actually a control system involved – if g>0–then your efforts to displace the mass would be resisted (in proportion to the size of g) and the mass would tend to move very little from the resting position r.Â

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BA: You get the same system behavior either way. They are mathematically equivalent.Â

RM: Then your “negative feedback” version of the equilibrium system is not a negative feedback model. If you put any negative feedback into the model (made g>0) then the mass would not behave the way it does in reality, even if g is very small. Â

BA: Either way, you have a system in which a disturbing force generates an opposing counterforce in such as way as to produce the observed behavior:

RM: Â The opposing counterforce has no feedback effect on the displacement that generates it. There is no feedback in the mass-spring system. “Equilibriium systems” are not feedback systems, period, amen. People who use these systems as examples of control systems are just trying to account for purposive behavior using causal models. They are like the person Bill described in B:CP (p. 8) who (“at a cybernetics conference yet” ) said that a drop of water sliding down a slightly inclined plane showed all the manifestations of purposeful behavior. The mass returning to its resting state after a displacement appears to behave purposefully in the same way. The people pushing “equilibrium” models of control (purpose) are just using a slightly more sophisticated version of this “raindrop analogy” of purpose. PCT explains (among other things) why these causal models are not control models.

Â

BA:Â The advantage of modeling these systems in terms of negative feedback is that you can then directly compare passive equilibrium systems to control systems.Â

RM: The disadvantage is that you give people the impression that equilibrium systems are negative feedback systems when they are not. Powers already did the modeling that compares equilibrium systems (Z systems) to control systems. He did it in B:CP and he did it quantitatively in the “Quantitative Analysis of Purposive Systems” paper in LCS I (p. 129). And he did a great job. Why muddy the waters?

Â

BA: The spring-mass system, for example, can be modeled as having a reference value equal to the undisturbed position of the mass and a  loop gain that is less than or equal to one but greater than zero. This approach makes explicit the differences between such a passive equilibrium system and an actual control system. The passive system’s “referenceâ€? value is fixed and equal to the undisturbed value; the control system’s reference can be adjusted.Â

RM:The difference between the mass spring system and a control system is that there is no feedback in the former – zero, nada, zilch. You don’t need to set g (the feedback constant) to anything other than 0 in order to model behavior of the mass spring system (as you do with a control system) because the output of the mass-spring system (the restoring force generated by the displacement) has no feedback effect on the cause of that force (the displacement). Indeed, if you model a mass spring system using g >0 the behavior of the model will not match the behavior of the real system.

Â

BA: The passive system’s loop gain cannot exceed 1.0

RM: The passive system has 0 loop gain because there is no loop! The restoring force (output) has no effect on the displacement of the mass (input) that caused it.Â

Â

BA: because its restorative energy comes from the disturbance itself; the control system’s loop gain can be extremely high because the energy it uses to generate the restorative force comes from an independent source. The passive equilibrium system will passively adjust to a continuous disturbance by moving to a new equilibrium value;

RM: I think you could write a great paper explaining why an equilibrium system is not in any way shape or form a kind of feedback system. You could show it mathematically and with modeling. And you could show, therefore, that if there is evidence that a variable (such as limb angle) is under control there must be a closed loop control system involved. And since equilibrium systems are not closed loop they cannot possibly explain the observed control. Now that would be a really nice contribution to PCT. How about it?

BestÂ

Rick


Richard S. Marken, Ph.D.
Author of  Doing Research on Purpose
Now available from Amazon or Barnes & Noble

From Bruce Abbott (2015.01.28.1700 EST)]

Rick Marken (2015.01.28.1120)–

Bruce Abbott (2014.01.28.0955 EST)

BA: Wrong. In the case of the spring, compressing the spring does mechanical work, stretching bonds within the material of the spring and generating the counterforce pushing back against the compressive force. In the case of the pendulum, raising the bob by swinging the pendulum at its pivot increases the potential energy of the bob (obtained from work done by the disturbance to raise the bob) and changes the angle from which gravity is pulling on the bob relative to the pendulum shaft. In both cases the result is a counterforce that acts against the disturbing force and increases with the amount of displacement.

RM: OK, I can buy that. But if you consider the displacement a disturbance then the counterforce that is generated is not very effective since the disturbance is fully effective. I think the important concept here is that the counterforce generated by the displacement is potential energy; it’s not energy that is actively working against the disturbance, as it would be if this were actually a feedback system. So even if it’s correct to say that the displacement (calling it a disturbance is kind of a misnomer since the position of the mass is not a controlled variable) generates the counter force, that counter force is not acting against the displacement as the displacement occurs, as it would in a closed loop control system. Equilibrium systems are not closed loop systems!

BA: Here’s the bottom line: You can model these systems using the appropriate formulas from physics (as you did) or explicitly as negative feedback systems (as I did for the spring example).

RM: I don’t believe you did model the mass spring system as a negative feedback system; if you had then you would have had to model it with two simultaneous equations:

F = k*(r-d) (1)

d = -g*F + Fp (2)

where F is the counter force, k is the spring constant, r is the resting position of the spring (assume it’s 0) , d is the actual position of the spring (when d<>0 there is displacement of the mass), g is the feedback constant (which I’ve made negative because you say this is a negative feedback system) and Fp is the force used to displace the mass (the equivalent of the disturbance variable, d, in PCT equations).

RM: Equation 1 is the “system” equation, relating input (position of the mass) to output (counter force generated by change in position). Equation 2 is the “environment” equation relating output (the counter force generated by your displacement of the mass) and environmental disturbances (your pull on the mass, Fp) to input.

RM: I think you will find that, in order to properly model the behavior of the mass on a spring, you would have to set the feedback constant, g, to 0, making the mass spring system an example of Powers’ Z system.That is, this “equilibrium system” is just a lineal causal system; there is no control involved at all. If there were actually a control system involved – if g>0–then your efforts to displace the mass would be resisted (in proportion to the size of g) and the mass would tend to move very little from the resting position r.

BA: You get the same system behavior either way. They are mathematically equivalent.

RM: Then your “negative feedback” version of the equilibrium system is not a negative feedback model. If you put any negative feedback into the model (made g>0) then the mass would not behave the way it does in reality, even if g is very small.

BA: Either way, you have a system in which a disturbing force generates an opposing counterforce in such as way as to produce the observed behavior:

RM: The opposing counterforce has no feedback effect on the displacement that generates it. There is no feedback in the mass-spring system. “Equilibriium systems” are not feedback systems, period, amen. People who use these systems as examples of control systems are just trying to account for purposive behavior using causal models. They are like the person Bill described in B:CP (p. 8) who (“at a cybernetics conference yet” ) said that a drop of water sliding down a slightly inclined plane showed all the manifestations of purposeful behavior. The mass returning to its resting state after a displacement appears to behave purposefully in the same way. The people pushing “equilibrium” models of control (purpose) are just using a slightly more sophisticated version of this “raindrop analogy” of purpose. PCT explains (among other things) why these causal models are not control models.

BA: If the opposing counterforce has no feedback effect, then what slows the rate of spring compression until it reaches zero? Magic?

BA: Look, here’s a model of the spring-mass-damper system entered into Simulink:

image00170.jpg

The first arrow after the summator circle at top left represents acceleration; the box it enters is an integrator. The arrow emerging from the integrator represents velocity. This enters a second integrator, whose output is position. The triangle labeled “Gain 1� actually holds the spring constant, which multiplies the position variable. The triangle labeled “Gain 2� holds the damping constant, which is multiplied by the velocity variable. The diagram does not show the mass, it divides both the velocity and position variables to produce forces. These forces feed back to affect the acceleration of the mass; note that the signs of both terms are negative. Thus you have, explicitly represented here, a negative feedback system.

In the simulation the spring’s relaxed position is at 0. The model run initializes the spring’s position at 1.0 (some force having been applied to put it there) and then releases the spring. The result is an oscillation that gradually dies out, as is expected from a damped spring-mass system.

If you aren’t convinced by that, then I don’t know what will. You can find the Simulink model being presented at

http://dasl.mem.drexel.edu/~rares/MassSpringDamper.htm

Still pushing the rock uphill,

Sisyphus

[From Rick Marken (2015.01.28.1640)]

image00170.jpg

···

Bruce Abbott (2015.01.28.1700 EST)–

BA: If the opposing counterforce has no feedback effect, then what slows the rate of spring compression until it reaches zero? Magic?

RM: What slows it is the counter force. Your mistake, I believe, is in thinking of this counter force as a “feedback” effect that counters a disturbance. In a feedback system the “countering” done by the system’s output is done while the disturbance is acting. In the mass- spring (and other “equilibrium” systems) the countering done by the system’s output (the counter force in the case of the mass-spring system) is done after the disturbance is removed (when you let go of the mass in the case of the mass-spring system, for example). Â

Â

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

F = k*(r-d) Â Â Â Â Â Â Â Â Â (1) Â

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

Â

d = -g*F + Fp       (2)Â

where g is the feedback constant and g <> 0. This equation recognizes the fact that while the value of d may be caused by the feedback effect of the counter force it is unquestionably caused by a downward or upward force, Fp, applied to the mass. If there is counter force feedback (g>0) and the feedback is negative (as shown by the minus sign) then the counter force will subtract from the applied force and d will be smaller than would be produced without the presumed feedback.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.

d = -1/k Fp   (3)

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

 BA: Look, here’s a model of the spring-mass-damper system entered into Simulink:

Â

Â

BA: The first arrow after the summator circle at top left represents acceleration; the box it enters is an integrator. The arrow emerging from the integrator represents velocity. This enters a second integrator, whose output is position. The triangle labeled “Gain 1� actually holds the spring constant, which multiplies the position variable. The triangle labeled “Gain 2� holds the damping constant, which is multiplied by the velocity variable. The diagram does not show the mass, it divides both the velocity and position variables to produce forces. These forces feed back to affect the acceleration of the mass; note that the signs of both terms are negative. Thus you have, explicitly represented here, a negative feedback system.

 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

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  effect on the the causes of that force.Â

RM: Another problem for me with the diagram is that it doesn’t show the effect of the disturbance on what is supposedly the “controlled variable” (the position of the mass). I think this is because equilibrium models are only interested in the stability that results after a disturbance has been removed. And that’s because equilibrium models don’t produce stability when there is a time-varying disturbance applied to the variable that is supposedly being stabilized. And that’s because there is no negative feedback in a an equilibrium model; there is no control.

Â

BA: In the simulation the spring’s relaxed position is at 0. The model run initializes the spring’s position at 1.0 (some force having been applied to put it there) and then releases the spring. The result is an oscillation that gradually dies out, as is expected from a damped spring-mass system.

 RM: Exactly. The model shows what happens after the disturbance (the force that was applied to move the mass to position 1.0). And what happens is exactly predicted by Newton’s lineal causal laws of motion. Â

 BA: If you aren’t convinced by that, then I don’t know what will.Â

RM: It’s way past convincing time for me. I have no idea why you think there is anything to these equilibrium models. But you are apparently as convinced of their value as I am convinced of their worthlessness. Indeed, worse than worthless.Â

RM I wish you wouldn’t waste your time and considerable talent on these frauds. But I can’t set your references for you. So there is really nothing I can do either. But, man, it would be great to have someone of your skill and intelligence working on PCT instead of the equilibrium stuff.

Best regards

Rick


Richard S. Marken, Ph.D.
Author of  Doing Research on Purpose
Now available from Amazon or Barnes & Noble

[From Bruce Abbott (2015.01.28.2210 EST)]

Rick Marken (2015.01.28.1640)–

Bruce Abbott (2015.01.28.1700 EST)–

BA: If the opposing counterforce has no feedback effect, then what slows the rate of spring compression until it reaches zero? Magic?

RM: What slows it is the counter force. Your mistake, I believe, is in thinking of this counter force as a “feedback” effect that counters a disturbance. In a feedback system the “countering” done by the system’s output is done while the disturbance is acting. In the mass- spring (and other “equilibrium” systems) the countering done by the system’s output (the counter force in the case of the mass-spring system) is done after the disturbance is removed (when you let go of the mass in the case of the mass-spring system, for example).

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:

F = k*(r-d) (1)

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

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.

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

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

d = -g*F + Fp (2)

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

where g is the feedback constant and g <> 0. This equation recognizes the fact that while the value of d may be caused by the feedback effect of the counter force it is unquestionably caused by a downward or upward force, Fp, applied to the mass. If there is counter force feedback (g>0) and the feedback is negative (as shown by the minus sign) then the counter force will subtract from the applied force and d will be smaller than would be produced without the presumed feedback.

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

d = -1/k Fp (3)

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.

BA:Â Well, garbage in, garbage out.

BA: Look, here’s a model of the spring-mass-damper system entered into Simulink:

image00231.jpg

BA: The first arrow after the summator circle at top left represents acceleration; the box it enters is an integrator. The arrow emerging from the integrator represents velocity. This enters a second integrator, whose output is position. The triangle labeled “Gain 1� actually holds the spring constant, which multiplies the position variable. The triangle labeled “Gain 2� holds the damping constant, which is multiplied by the velocity variable. The diagram does not show the mass, it divides both the velocity and position variables to produce forces. These forces feed back to affect the acceleration of the mass; note that the signs of both terms are negative. Thus you have, explicitly represented here, a negative feedback system.

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.

BA: Now this is getting downright bizarre. 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: 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 effect on the the causes of that force.

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

RM: Another problem for me with the diagram is that it doesn’t show the effect of the disturbance on what is supposedly the “controlled variable” (the position of the mass). I think this is because equilibrium models are only interested in the stability that results after a disturbance has been removed. And that’s because equilibrium models don’t produce stability when there is a time-varying disturbance applied to the variable that is supposedly being stabilized. And that’s because there is no negative feedback in a an equilibrium model; there is no control.

The Simulink model of the mass-spring-damper system was set up to demonstrate what happens after an impulse disturbance that produced a deviation of 1 unit from the spring’s resting position. If the disturbance were to be applied continually, beginning with the spring at its resting position, the same model would show the spring compressing and the position of the mass changing until it reached a steady value dependent on the size and direction of the disturbing force, minus the counterforce being generated in the spring, the latter being determined by the deflection of the mass times the spring constant.

I don’t know why you added “there is no control� to the end of your last sentence. It suggests that I am asserting that passive equilibrium systems are control systems, which I have taken pains to make clear they are not. (They do, however, reduce the effects of disturbances to some degree.)

BA: In the simulation the spring’s relaxed position is at 0. The model run initializes the spring’s position at 1.0 (some force having been applied to put it there) and then releases the spring. The result is an oscillation that gradually dies out, as is expected from a damped spring-mass system.

RM: Exactly. The model shows what happens after the disturbance (the force that was applied to move the mass to position 1.0). And what happens is exactly predicted by Newton’s lineal causal laws of motion.

BA: If you aren’t convinced by that, then I don’t know what will.

RM: It’s way past convincing time for me. I have no idea why you think there is anything to these equilibrium models. But you are apparently as convinced of their value as I am convinced of their worthlessness. Indeed, worse than worthless.

BA: Well, here’s my last try at convincing you. I discovered one last arrow in my quiver. Here it is:

procedure TMainForm.RunModel;

begin

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

 Accel := (Force + ForceDist - DVel - SPos)/M;

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

 // New position = old position + velocitydt + 1/2 accelerationdt-squared

 Pos := Pos + (Vel + 0.5Acceldt)*dt;

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

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

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

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

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

end;

This is a control system acting on a mass-spring-damper system.  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. Notice the label over the Accel variable:

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

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

Time to concede, Rick.

Bruce

2015,01
Angus Jenkinson

BA: Here’s the bottom line: You can model these systems using the appropriate formulas from physics (as you did) or explicitly as negative feedback systems (as I did for the spring example). You get the same system behavior either way. They are mathematically equivalent.

AJ Interesting remark
Older pre modernist systems used the phrase saving the appearances. It referred to the ability of a model to reproduce phenomena and predict them. Thus Ptolemy saved the appearances better than Copernicus. They never believed that the model was reality only a device for practical purposes, which over time was forgotten. The model was reified. So the proposition here is that two descriptions or calculation modes (models) produce the same answer. If this is indeed true what might be a conclusion?

[Martin Taylor 2014.01.29.13.47]

2015,01
Angus Jenkinson

BA: Here’s the bottom line: You can model these systems using the appropriate formulas from physics (as you did) or explicitly as negative feedback systems (as I did for the spring example). You get the same system behavior either way. They are mathematically equivalent.

AJ Interesting remark
Older pre modernist systems used the phrase saving the appearances. It referred to the ability of a model to reproduce phenomena and predict them. Thus Ptolemy saved the appearances better than Copernicus. They never believed that the model was reality only a device for practical purposes, which over time was forgotten. The model was reified. So the proposition here is that two descriptions or calculation modes (models) produce the same answer. If this is indeed true what might be a conclusion?

The difference between Ptolemy-Compernicus and what Bruce is saying is that in P-C, the model of the hidden behaviour of the planets was different. To advance the science, a proponent of the Ptolemaic model would have had to offer a mechanism of producing the epicycles that did not rely on a circular argument. Similarly for a Copernican enthusiast, though the C-ist would need fewer parameters in the mechanism (and much later, Newton showed that only two were required to account for all Kepler's refinement of Copernicus's model).

In the case of Bruce-Rick, the model of the hidden behaviour is identical. The only difference I can see is in the words used to describe the same mathematical model. Rick is exclusionary, and rejects the word "feedback", whereas Bruce notes that it's exactly the same model whichever wording you use, and that to use the word "feedback" allows the spring model to take its natural place in a hierarchy of feedback systems in much the way the non-number "zero" takes its place in the series of counting numbers. If you count the number of cows in a field and the field is empty of mammals, it's perfectly natural for us now to say there are zero cows. A while back, the answer might have been "How can I count them? There aren't any there!". I think the issue is not unlike that.

Martin

PS. Angus, it helps us to back-reference an old post if you start with an ID line, which has conventionally included the poster's name and a time-and-date stamp in International Year-month-day-hour-minute format. But it really doesn't much matter what you use as an ID, provided we can seek back in the archive for it and be sure we have the proper message.

···

On 2015/01/29 1:32 PM, Angus Jenkinson (angus@angusjenkinson.com via csgnet Mailing List) wrote:

[From Bruce Abbott (2015.01.29.1405 EST)]

2015,01
Angus Jenkinson

BA: Here’s the bottom line: You can model these systems using the appropriate formulas from physics (as you did) or explicitly as negative feedback systems (as I did for the spring example). You get the same system behavior either way. They are mathematically equivalent.

AJ Interesting remark
Older pre modernist systems used the phrase saving the appearances. It referred to the ability of a model to reproduce phenomena and predict them. Thus Ptolemy saved the appearances better than Copernicus. They never believed that the model was reality only a device for practical purposes, which over time was forgotten. The model was reified. So the proposition here is that two descriptions or calculation modes (models) produce the same answer. If this is indeed true what might be a conclusion?

You're mixing apple and oranges. Ptolemy and Copernicus offered two different models of the solar system that were compatible with the available astronomical observations. Ptolemy's system actually predicted the motions of the planets and moon more accurately, but at the expense of complexity.

The two approaches I referred to employ the same model but compute the solution differently. The physics formula-based approach solves the problem using a system of differential equations; the Simulink (modeling) approach does it numerically.

Bruce

2015,1,29
Angus Jenkinson

Martin
An interesting answer, thank you.
Btw I suppose you know Copernicus retained epicycles and a comparable number? His variation was ideological rather than scientific.

[From Rick Marken (2015.01.29.2210)]

···

 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

Â

F = k*(r-d) Â Â Â Â Â Â Â Â Â (1)

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

Â

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.

Â

BA: r-p is the deviation. The cause is some force that is acting to compress the spring. I’ve referred to this force as a disturbance. 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       (2)Â

Â

BA: Now you are defining d as the difference between two forces. 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           (1)Â

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

x = -g*F + Fp       (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

Â

d = -1/k Fp   (3)

Â

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

Â

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! 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

 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

Â

BA: Now this is getting downright bizarre. 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

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

Â

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  effect on the the causes of that force.Â

Â

BA: You are asserting that compression of a spring by a force does not simultaneously produce a counterforce as the spring pushes back. 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. 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:  I don’t know why you added “there is no control� to the end of your last sentence. It suggests that I am asserting that passive equilibrium systems are control systems, which I have taken pains to make clear they are not. (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: Well, here’s my last try at convincing you. I discovered one last arrow in my quiver. Here it is:

Â

procedure TMainForm.RunModel;

begin

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

 Accel := (Force + ForceDist - DVel - SPos)/M;

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

 // New position = old position + velocitydt + 1/2 accelerationdt-squared

 Pos := Pos + (Vel + 0.5Acceldt)*dt;

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

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

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

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

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

end;

Â

This is a control system acting on a mass-spring-damper system.  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. Notice the label over the Accel variable:

Â

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

Â

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

Â

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

2015,1,30, Angus Jenkinson

Martin said: The difference between Ptolemy-Compernicus and what Bruce is saying is

that in P-C, the model of the hidden behaviour of the planets was

different. To advance the science, a proponent of the Ptolemaic model

would have had to offer a mechanism of producing the epicycles that did

not rely on a circular argument. Similarly for a Copernican enthusiast,

though the C-ist would need fewer parameters in the mechanism (and much

later, Newton showed that only two were required to account for all

Kepler’s refinement of Copernicus’s model).

In the case of Bruce-Rick, the model of the hidden behaviour is

identical.

AJ: Martin, Your comments on the two perspectives are interesting. However, I have a rather different POV on the P-C distinction and the point I was making. Firstly, I was making a distinction between saving the appearances and a model that is reified into “what is actually going on”. In the scientific view of saving the appearances, it was never assumed that this was what was actually going on in physical reality. Today astronomers still use the inside of "a Celestial Globe” to portray the location of planets and so forth as was done in the past and at least at this level neither originally meant it to be taken literally. So my curiosity related to two (at the time anyway) seemingly equivalent descriptions using different paradigms

And secondly the distinction between Ptolemy and Copernicu
s seems not to be as you say. There is more mythology and misinformaiton on this subject than can be easily imagined. the physicist Fitzpatrick did some fascinating reworking to put Ptolemy into modern language and mathematical method and corrected many assumptions. Most of the quoted remarks are based on Copernicus’s a) first notes not his final published version, the C version had an equivalent or greater number of epicycles as P (looks like Copernicus had 48 to P’s 40 b) C’s objection to equants, which were actually a device that reproduce ellipses. C objected because he believed in uniform motion, a Platonic principle. He reverts to abstract principles while Ptolemy follows the phenomena. Moreover his movement of the sun to centre was religious conviction not a scientific one, he was a Rosicrucian and the sun represented Christ, the sol invicta.

Fitzpatrick comments: "Ptolemy’s aim in the Almagest is to construct a kinematic model o
f the solar system, as seen from the earth. In other words, the Almagest outlines a relatively simple geometric model which describes the apparent motions of the sun, moon, and planets, relative to the earth, but does not attempt to explain why these motions occur.”

Fred Hoyle commented, “Today, we cannot say that the Copernican theory is “right” and the Ptolemaic theory “wrong” in any meaningful physical sense. The two theories, when improved by adding terms involving the square of higher powers of the eccentricities of the planetary orbits, are physically equivalent to one another.” [“Nicolaus Copernicus,” Harper & Row, Publishers, 1973, page 88]

But this is not to the point of our main conversations

2015, 1,30 Angus Jenkinson

Bruce said: You’re mixing apple and oranges. Ptolemy and Copernicus offered two different models of the solar system that were compatible with the available astronomical observations. Ptolemy’s system actually predicted the motions of the planets and moon more accurately, but at the expense of complexity.

The two approaches I referred to employ the same model but compute the solution differently. The physics formula-based approach solves the problem using a
system of differential equations; the Simulink (modeling) approach does it numerically.

Cross-referencing to my other reply, P’s system was not more complicated. That was a myth perpetuated by an historical cultural tradition (still maintained) that needed to elevate C and his critique over P. Modern analysis shows that they were equivalent and C was no more elegant.

As to apple and oranges, PC and RB may have more in common than you present. I don’t wish to ascribe descriptions of your efforts, which I am sure you understand better than I. However, as far as I could see at the time of commenting, both offer " two different models of th
e … system that were compatible with the available… observations but compute the solution differently.”

This is not to say that the difference does not matter.

From Angus Jenkinson, 2015, 1,30,11:39GMT

The technical discussion between Rick and Bruce is very interesting. I would like to insert another parameter. It is one that interests me more because I am primarily working in the organic realm and human behaviours. What most interests me about PCT is that it involves viable organisms actively managing their behaviours to maintain intended paths to goals (by controlling perceptions). There is nothing in a spring or a pendulum like this. Moreover, while causality operates (I think) more or less as the traditional view of causality, at least in immediate proximity to the effect, in PCT in the human realm an utterly different process occurs of actively cancelling perturbations (external causal factors). PCT is not being applied to springs.

[From Bruce Abbott (2015.01.29.0855 EST)]

Rick Marken (2015.01.29.2210)–

Bruce Abbott (2015.01.28.2210 EST)

BA: Now this is getting downright bizarre. 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.

BA: Cutting to the chase . . .

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.

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:

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

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.

BA: I would welcome Richard’s input here, but alas, there is never a professional mathematician around when you need one . . . (;->

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 effect on the the causes of that force.

BA: You are asserting that compression of a spring by a force does not simultaneously produce a counterforce as the spring pushes back. 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. 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).

BA: In my world, feedback occurs when changing a variable produces effects that act on the same variable. When the effect acts in a direction that opposes the original change, it is called negative feedback.

BA: In the case of a spring, a force acting to compress (or stretch, for that matter) a mass-spring system accelerates the mass, thus increasing its velocity and thus changing its position. As the spring compresses, a counterforce develops that is proportional to the change in the length of the spring (and thus to the change in position of the mass). This force acts on the mass in a direction opposite to that of the compressive force, opposing the acceleration of the mass.  In other words, the change in position of the mass feeds back to oppose that change in position. This meets the formal definition of negative feedback. Â

BA: So if you’re going to assert that there is no feedback here, you are going to have to prove that negative feedback is not as I have defined it. So, what is your alternative definition of negative feedback?

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: The mass would continue to accelerate away from its resting position if where were no “countering� being done. The fact that the system comes to rest at a given amount of displacement of the mass demonstrates that there is a counterforce and that it is doing enough “countering� to bring the mass to rest despite the continuation of the disturbing force.

BA: I don’t know why you added “there is no control� to the end of your last sentence. It suggests that I am asserting that passive equilibrium systems are control systems, which I have taken pains to make clear they are not. (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. A good control system reduces the effects of disturbances almost completely; a poor control system reduces the effects of disturbances "to some degree’.

BA: Then given your definition of control, passive equilibrium systems exert some degree of control – their negativve feedback reduces the change that disturbances produce. But I say that passive equilibrium systems are not control systems as I define the term. Control systems employ an external source of energy to generate their opposing actions; for this reason their loop gains can be much higher than 1.0. Equilibrium systems extract energy from the disturbance itself and therefore cannot have a loop gain exceeding 1.0 (and most likely will be less due to energy losses such as those due to friction). Control systems have reference values than can in principle be set to different values and the control system will then act to make the input value approximate the reference value. Equilibrium systems simply settle to values at which the various forces come into balance (equilibrium).

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, 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.

BA: Well, here’s my last try at convincing you. I discovered one last arrow in my quiver. Here it is:

procedure TMainForm.RunModel;

begin

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

Accel := (Force + ForceDist - DVel - SPos)/M;

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

// New position = old position + velocitydt + 1/2 accelerationdt-squared

Pos := Pos + (Vel + 0.5Acceldt)*dt;

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

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

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

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

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

end;

This is a control system acting on a mass-spring-damper system. 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. Notice the label over the Accel variable:

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

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

BA: No, “spring feedback� is S*Pos, where S is the spring constant and Pos is the deviation of the mass from its resting position. This computes the level of counterforce generated by the spring. It is subtracted because the feedback is negative. “Velocity feedback� is the force being generated by the damper, which is proportional to the velocity of the mass. (D is the damping coefficient setting the amount of counterforce generated per unit of velocity. This force is also being subtracted because it constitutes another source of negative feedback. These two sources of negative feedback are what are represented in the Simulink diagram as lines emerging from the triangular “gain� symbols and entering, with negative signs, the circular summation icon.

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!

BA: Your statement above has clarified for me why you are so adamantly opposed to believing the fact that passive equilibrium systems help to stabilize variables through negative feedback. But this is not an attempt to use a “causal� model (by which I take you to mean “lineal causal model�) to account for purposeful behavior. Equilibrium systems do not exhibit purpose, they just come to whatever equilibrium values emerge from the interaction of their variables. Purposeful behavior involves an active intervention (energy drawn from an independent source) that pushes the controlled variable toward an intended  (reference) value.

Bruce

[From Rick Marken (2015.01.30.1230)]

···

Bruce Abbott (2015.01.29.0855 EST)–

 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  effect on the the causes of that force.Â

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

RM: No, I’m saying that there is no counter force going from F to the cause of F, which is x (displacement of the mass). Hooke’s law says

F = kx

So there is a causal path from x to F; but there is no simultaneous causal path from F back to x, ie, Hooke law doesn’t imply that x = -g F; that would be negative feedback. By calling F a “counter force” I believe you are verbally deceiving yourself (and deceiving your audience) into thinking of F as negative feedback. It ain’t.

 BA: In my world, feedback occurs when changing a variable produces effects that act on the same variable. When the effect acts in a direction that opposes the original change, it is called negative feedback.

RM: Same in my world. And as I noted above, there is no feedback in the mass spring system. Changing the position of the mass (x) produces an effect (F) but that effect does not “feed back” and oppose that change.Â

Â

 BA: In the case of a spring, a force acting to compress (or stretch, for that matter) a mass-spring system accelerates the mass, thus increasing its velocity and thus changing its position. As the spring compresses, a counterforce develops that is proportional to the change in the length of the spring (and thus to the change in position of the mass). This force acts on the mass in a direction opposite to that of the compressive force, opposing the acceleration of the mass.Â

RM: I think you are trying to describe Newton’s 3rd law of motion as negative feedback. What we know about the mass-spring system is that a force (Fp) that moves a mass by an amount x from it’s resting position generates a “counter” or restoring force (F) in the spring  that is proportional, by the spring constant k, to x. That’s Hooke’s law:

F = kx       (1)

This law implies that the amount that the force Fp – the force that pulls or pushes the mass – can move a mass from its resting position is

x = -1/k(Fp) Â Â Â (2)

the minus sign indicating that Fp is applied in a direction opposite to F. That is, if Fp is a downward pull F is an upward pull; if Fp is an upward push F is a downward push.

Substituting equation 2 into equation 1 we get

F =- k(1/k) Fp  and

F = -Fp

which is simply the third law of motion: for every action (Fp) Â there is an equal and opposite reaction (F). This is not negative feedback.

Â

BA: In other words, the change in position of the mass feeds back to oppose that change in position. This meets the formal definition of negative feedback. Â

RM: Forces acting in opposite directions are not necessarily opposing one another. They are opposing each other in a control system but they are simply acting in opposite directions in a passive “equilibrium” system like that of a mass on a spring.

Â

 BA: So if you’re going to assert that there is no feedback here, you are going to have to prove that negative feedback is not as I have defined it.Â

RM: I think I just did. F, the “counter force” generated by pulling the mass from its resting position (creating deviation x), is simply the opposite of the force, Fp, that produced the deviation; there is no “opposition” of F to Fp; they are the same force operating in opposite directions, per Sir Isaac.Â

Â

RM: So, what is your alternative definition of negative feedback?

RM: I’ve already given it. Negative feedback exists when the cause of an output is affected (negatively) by that output. So if F is considered an output of the mass spring system, there is negative feedback if

F = kx and  x = -gF

Â

BA:Â Then given your definition of control, passive equilibrium systems exert some degree of control – theirr negative feedback reduces the change that disturbances produce.Â

RM: Yes, they would be control systems if they actually did reduce the effect of disturbance on the controlled variable. But they don’t so they aren’t.Â

Â

BA: But I say that passive equilibrium systems are not control systems as I define the term.Â

RM: Yes! They are not control systems because they don’t control.

Â

Control systems employ an external source of energy to generate their opposing actions; for this reason their loop gains can be much higher than 1.0. Equilibrium systems extract energy from the disturbance itself and therefore cannot have a loop gain exceeding 1.0 (and most likely will be less due to energy losses such as those due to friction).Â

RM: Equilibrium systems cannot have a loop gain greater than 1 because there is no loop in these systems.** **The loop gain of passive equilibrium systems is 0; they are Z systems.Â

BA: Control systems have reference values than can in principle be set to different values and the control system will then act to make the input value approximate the reference value. Equilibrium systems simply settle to values at which the various forces come into balance (equilibrium).

RM: Yes, they have no reference values at all because they are not closed loop systems; there is no controlled input to specify with a variable or a fixed reference.Â

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

Â

BA: No, “spring feedback� is S*Pos,

RM: Ah, there’s Hooke’s law. Yes, if Bill called SPos “spring feedback” then he was being careless in his terminology. SPos is simply the restoring force generated by the deviation of position of the spring (muscle) from it’s resting position, which I presume is what Pos is.Â

 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

Â

BA:Â Your statement above has clarified for me why you are so adamantly opposed to believing the fact that passive equilibrium systems help to stabilize variables through negative feedback.Â

RM: If equilibrium systems are negative feedback systems then they are control systems. In which case they don’t help stabilize; they stabilize. I think of “stabilizing” as being the same as “controlling”. If there is a difference between stabilizing and controlling you will have to explain how one can determine the difference between them by observation. So the reason I am so adamantly opposed to the idea that “passive equilibrium systems” help to stabilize variables through negative feedback" is that I can’t see it as anything other than an attempt to show that control (stabilization) can be exerted by causal systems.Â

BA: But this is not an attempt to use a “causalâ€? model (by which I take you to mean “lineal causal modelâ€?) to account for purposeful behavior. Equilibrium systems do not exhibit purpose, they just come to whatever equilibrium values emerge from the interaction of their variables.Â

RM: I see a big difference between “coming to whatever equilibrium state emerges from the interaction of variables” and “stabilizing variables”. “Stabilizing” implies active opposition to disturbance. “Coming to whatever equilibrium state emerges from the interaction of variables” implies no opposition to disturbance; disturbances are just one of the variables that determines the equilibrium state. Â Â

RM: I think the main thing you have to do to convince me that equilibrium systems “help stabilize variables through negative feedback” is explain the difference between stabilization and control. This explanation should include a description of a method I can use to distinguish, by observation, the difference between a variable that is being stabilized and one that is being controlled. Once I know how to tell the difference between stabilization and control you can start explaining why the negative feedback in a passive equilibrium system (which I don’t believe exists) produces stabilization (but not control) and why the negative feedback in a control system produces control (but not stabilization?).Â

Best

Rick


Richard S. Marken, Ph.D.
Author of  Doing Research on Purpose
Now available from Amazon or Barnes & Noble

[Martin Taylor 2011.01.30.15.48]

What an EXTRAORDINARY claim!!!!!!!!

All negative feedback systems are control systems, indeed!
And this claim is the rationale for saying that a spring is not a
negative feedback system. Bruce said an equilibrium system is not a
control system, and you agreed, so your claim that any negative
feedback system is a control system PROVES that there’s no negative
feedback in the spring. Ever heard of circular argument?
As the English professor said: “Two negatives make a positive, but
two positives don’t make a negative”.
To which a student at the back of the room said: “Yeah, Yeah”.
Wow!
The mind boggles indeed. Rick never sat in a chair or lay on the a
bed!
Well, I guess when one is in Wonderland, Humpty-Dumpty has some
authority.
Martin

···

[From Rick Marken (2015.01.30.1230)]

                Bruce

Abbott (2015.01.29.0855 EST)–

          RM: If equilibrium systems are negative feedback

systems then they are control systems.

          In which case they don't help stabilize; they

stabilize. I think of “stabilizing” as being the same as
“controlling”.