# Mass-Spring-Damper Simulation

[From Bruce Abbott (2015.02.05.2120 EST)]

Given the recent debate here about the behavior of passive equilibrium systems, I thought it would save a lot of “blah blah” argumentation to present a working simulation of such a system – in this case a mass-spring-damper system. I’ve created such a simulation (it’s a Windows App but should run on Macs that can spoof the Windows operating system) and posted it to my “Perceptual Control System Demos” Google website. If you would like a copy just follow this link (https://sites.google.com/site/perceptualcontroldemos/home/other-demos ) and click the “download” link under MassSpringDamperModel.zip. The zip file includes the executable (look for the .exe file extension in the Debug/Win32 subdirectory) and all necessary source code. You can copy the .exe file to your desktop or anywhere else you like as it does not require any other files to run. Then just double-click on the program’s icon.

The mass-spring-damper model is found in the Unit1.pas file, which is a text file so you can read it even if you don’t have the Delphi IDE. The model itself is contained in procedure TForm1.RunSim, as shown below:

procedure TForm1.RunSim;

// Computes the variables of the model each iteration

begin

Acc := (Force - dVel - sPos)/Mass; // net acceleration

Pos := Pos + (Vel + 0.5Accdt)*dt; // integrate position

Vel := Vel + Acc*dt; // integrate velocity

Time := Time + dt;

if Time >= ForceDur then Force := 0.0;

end;

The relevant variables and parameters are initialize in another procedure. During a model run, the RunSim procedure executes over and over in a loop, generating the behavioral dynamics of the system.

The first line computes the net acceleration due to forces acting on the mass: Force is an external force that compresses the spring; dVel gives the reaction force generated by the damper (d is the damping coefficient and Vel the velocity of the mass); sPos gives the reaction force generated by the spring, where s is the spring constant and Pos is the position of the mass (relative to its resting state, which is defined as zero. These forces are summed to give the net force and divided by the mass to give the acceleration of the mass as given by Newton’s force = Mass*Acc, solved for acceleration (i.e., Acc = force/mass).

The second line computes the new position from the previous one plus the change in position during the simulation time-step, dt. This change is found (approximately) by taking the previous velocity and adding one-half of the acceleration, multiplied by the time-step. The resulting new velocity is then multiplied by the time-step to yield the change in position.

The third line takes the old velocity and adds the change in velocity during the time-step, computed by multiplying the acceleration times the time-step.

The fourth line keeps track of the current time since the simulation began. This time is compared in the fifth line with the time stored in ForceDur (length of time the external force is allowed to continue acting on the mass); when the force has been acting for the allotted time, it is removed.

Here is a screen shot showing the simulation after it has been running long enough to complete the graphs shown in the bottom panel: In this run, the external force pushed against the spring for seven seconds. The three graphs shown at bottom represent the position, velocity, and acceleration of the mass, respectively, over the course of several seconds. The sketch of the mass-spring-damper arrangement above the graphs shows action live.

Note especially what happens to the position of the mass. At time zero the force is suddenly applied to the mass, accelerating it to the left, compressing the spring and moving the damper. In the graph, up indicates movement to the left. Initially there’s a bit of oscillation but this soon settles down to a new position well to the left of the resting position. This lays to rest any idea that the effect of the “negative feedback” is evident only after the external force is removed. If there were no “negative feedback, the impressed force would continue to accelerate the mass so long as the force persisted, but as you can see, this is not what happens. The system develops some degree of resistance to being disturbed from its resting state – enough resistance to bring the mass to a halt at a new position.

At the seven second mark the impressed force suddenly disappears, leaving the counterforces (“negative feedback”) to accelerate the mass toward the initial resting point. After some oscillation, the damping forces (which are proportional to velocity) convert the remaining energy of the system to heat and the system returns to its initial resting position.

So here we have a system that resists disturbances (but does not resist them as strongly as a control system does) and returns to its initial equilibrium position once the disturbance is removed.

As the picture show, you can play around with various parameters (spring coefficient, damping coefficient, etc.) to see how they affect the way the system behaves. Be aware, however, that these parameters can be adjusted only while the simulation is not in the process of plotting the graphs. The system continues to run even after the graphs are complete, so you will need to press the “RESET” button before you can adjust the parameters. Then press the “START” button again to begin a new run with the altered parameters.

[From Rick Marken (2015.02.05.2300)] ···

Bruce Abbott (2015.02.05.2120 EST)–

Â

BA: Given the recent debate here about the behavior of passive equilibrium systems,Â I thought it would save a lot of âblah blahâ? argumentation to present a working simulation of such a system  in this case a mass-spring-damper syystem.

RM: Beautiful job, as usual, Bruce! This will illustrate the point I was going to make perfectly.

Â BA:Â Here is a screen shot showing the simulation after it has been running long enough to complete the graphs shown in the bottom panel:

Â

Â

BA: In this run, the external force pushed against the spring for seven seconds.Â The three graphs shown at bottom represent the position, velocity, and acceleration of the mass, respectively, over the course of several seconds.Â The sketch of the mass-spring-damper arrangement above the graphs shows action live.

Â RM: Very nice job.

BA: Note especially what happens to the position of the mass.Â

RM: Yes, that is very important.

Â

BA: At time zero the force is suddenly applied to the mass, accelerating it to the left, compressing the spring and moving the damper.Â In the graph, up indicates movement to the left.Â Initially thereâs a bit of oscillation but this soon settles down to a new position well to the left of the resting position.Â

RM: Yes, the variable called Force is the disturbance to the position, Pos, of the mass. What is clear from the graph is that this disturbance is completely effective. The disturbance Force moves the mass to the position expected it would be expected to move to if there were no negative feedback at all.Â

RM: The expected effect of the disturbance Force is to move the mass to a position that is proportional to the reciprocal of the spring constant times the Force. That is, the expected value of Pos after the application of the disturbing Force is 1/s(Force). And that seems to be exactly what’s shown in your simulation This is easiest to see if you set the damping to 10 to cut down the transient overshoot and oscillation that is seen when the disturbance Force is applied (and removed) abruptly. When you run the simulation with the spring constant s = 100 Pos ends up about 5 times higher on the Position graph than it does when the spring constant s= 500. Of course, when the disturbance Force goes to 0 the mass returns to the 0 position regardless of the spring constant since 1/s * 0 is zero no matter what the value of s.

RM: It would be nice if you could put a line one the chart indicating the expected steady state position of the mass after the application of a non-zero disturbance Force. I think you will find that the position of the mass will end up right on that line after a brief period of damped oscillation.Â

BA: This lays to rest any idea that the effect of the ânegative feedbackâ? is evident only after the external force is removed.Â

RM: Actually, the simulation shows that negative feedback isn’t evident at all. If there were any negative feedback – any resistance at all to the effect of the disturbance Force – then the disturbance force (Force) would have less effect on the position of the mass than expected from physics. But the disturbance Force has exactly the expected effect indicating no negative feedback at all; the mass-spring system is a Z-system (open-loop causal system).Â

Â

BA: If there were no ânegative feedback, the impressed force would continue to accelerate the mass so long as the force persisted,

RM: Actually, this would only happen if the spring and damper were orbiting the earth (instead of being attached to a solid wall) and the force were applied by thruster engines.Â

Â

but as you can see, this is not what happens.Â The system develops some degree of resistance to being disturbed from its resting state  enough resistance to brinng the mass to a halt at a new position.

RM: Where do you see resistance to disturbance? As far as I can see, the simulation shows that the disturbance (Force) is completely effective.Â

Â

BA: At the seven second mark the impressed force suddenly disappears, leaving the counterforces (ânegative feedbackâ?) to accelerate the mass toward the initial resting point.Â After some oscillation, the damping forces (which are proportional to velocity) convert the remaining energy of the system to heat and the system returns to its initial resting position.

RM: Again, that is exactly the expected effect on the position (Pos) of the mass if the disturbance Force goes to 0. Â

Â

Â BA: So here we have a system that resists disturbances (but does not resist them as strongly as a control system does) and returns to its initial equilibrium position once the disturbance is removed.

RM: I see a beautiful demonstration of the fact that a mass-spring system does not put up any resistance to the effect of a disturbance (Force) on a variable (Pos) affected by that disturbance. Â

BestÂ

Rick

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

[From Bruce Abbott (2015.02.06.0900 EST)]

Rick Marken (2015.02.05.2300)]

Bruce Abbott (2015.02.05.2120 EST)–

BA: Given the recent debate here about the behavior of passive equilibrium systems, I thought it would save a lot of â€œblah blahâ€? argumentation to present a working simulation of such a system – in this case a mass-sspring-damper system.

RM: Beautiful job, as usual, Bruce! This will illustrate the point I was going to make perfectly.

BA: Here is a screen shot showing the simulation after it has been running long enough to complete the graphs shown in the bottom panel: BA: In this run, the external force pushed against the spring for seven seconds. The three graphs shown at bottom represent the position, velocity, and acceleration of the mass, respectively, over the course of several seconds. The sketch of the mass-spring-damper arrangement above the graphs shows action live.

RM: Very nice job.

BA: Thanks!

BA: Note especially what happens to the position of the mass.

RM: Yes, that is very important.

BA: At time zero the force is suddenly applied to the mass, accelerating it to the left, compressing the spring and moving the damper. In the graph, up indicates movement to the left. Initially thereâ€™s a bit of oscillation but this soon settles down to a new position well to the left of the resting position.

RM: Yes, the variable called Force is the disturbance to the position, Pos, of the mass. What is clear from the graph is that this disturbance is completely effective. The disturbance Force moves the mass to the position expected it would be expected to move to if there were no negative feedback at all.

BA: Iâ€™ve added bold font to call attention to the relevant assertion.Â You are starting to sound a bit like George Orwellâ€™s 1984, in which a dictatorial regime redefines terms to mean what the regime wants them to mean:Â â€œWar is peace, freedom is slavery, ignorance is strength.â€?

BA: For you, â€œcompletely ineffectiveâ€? means that the system comes to rest (under continued applied force) exactly where one would expect it to come to rest based on the laws of physics.Â If you apply this definition to the operation of a control system that applies force to a CV in order to control position, then by the same token you must conclude that the control system also comes to rest exactly where one expects it to based on the laws of physics.Â That is the point at which the counterforce supplied by the control system equals the disturbing force acting on the controlled variable.Â Just as in the spring-mass-damper system, the counterforce supplied by the control system will limit how much change occurs to position.Â The only difference between the control system and the equilibrium system in this case is in the amount of change in position that the applied force will produce.Â

BA: The equilibrium system being subjected to a continued disturbing force will see the mass move to the point at which the reaction force (as described by Hookâ€™s Law) equals the disturbing force.Â The spring must be compressed a certain distance in order to produce this equal reaction force, and that is the position at which the system at rest will maintain.

BA: Assume a proportional control system for positioning a mass (no springs here).Â This control system, being subjected to a continued disturbing force, will see the mass move to the point at which the counterforce produced by the control system exactly equals the applied force.Â To generate this counterforce the disturbing force must move the mass far enough from its reference position that the error between reference and position, acting on the output function, produces a force equal to the disturbing force.Â The amount of distance required to achieve this equilibrium of forces will depend on the loop gain of the control system.

BA:Â From this comparison it can be seen that the proper definition of â€œineffective in resisting disturbanceâ€? in these two systems is NOT that the position of the mass moves exactly as one would expect given the applicable laws of physics (because the equilibrium system and the control system both do that).Â The proper definition of ineffective would be that the disturbance is unopposed by the system.Â So, the argument you are presenting here is based on adopting a faulty definition of â€œineffective.â€?Â When you adopt the correct definition, it can be seen that both systems are â€œeffective in resisting the disturbanceâ€? but to different degrees.Â As Iâ€™ve pointed out ad nauseum, the difference is due to the fact that the equilibrium system gets its restorative force from the energy supplied by the disturbance itself, whereas the control system has an independent source of energy to power its opposing force, and can generate a far stronger opposition to the extent that its loop gain well exceeds 1.0.

RM: The expected effect of the disturbance Force is to move the mass to a position that is proportional to the reciprocal of the spring constant times the Force. That is, the expected value of Pos after the application of the disturbing Force is 1/s(Force). And that seems to be exactly what’s shown in your simulation This is easiest to see if you set the damping to 10 to cut down the transient overshoot and oscillation that is seen when the disturbance Force is applied (and removed) abruptly. When you run the simulation with the spring constant s = 100 Pos ends up about 5 times higher on the Position graph than it does when the spring constant s= 500. Of course, when the disturbance Force goes to 0 the mass returns to the 0 position regardless of the spring constant since 1/s * 0 is zero no matter what the value of s.

BA:Â Yes, itâ€™s exactly what you expect in this system owing to the reaction force that develops as the spring is compressed.Â The reaction force, acting against the effect of disturbing force increases until it equals the disturbing force.Â Thereâ€™s a name for this kind of influence.Â Itâ€™s called â€œnegative feedback.â€?

RM: It would be nice if you could put a line one the chart indicating the expected steady state position of the mass after the application of a non-zero disturbance Force. I think you will find that the position of the mass will end up right on that line after a brief period of damped oscillation.

BA: This lays to rest any idea that the effect of the â€œnegative feedbackâ€? is evident only after the external force is removed.

RM: Actually, the simulation shows that negative feedback isn’t evident at all. If there were any negative feedback – any resistance at all to the effect of the disturbance Force – then the disturbance force (Force) would have less effect on the position of the mass than expected from physics. But the disturbance Force has exactly the expected effect indicating no negative feedback at all; the mass-spring system is a Z-system (open-loop causal system).

BA:Â So if itâ€™s â€œexpected from physics,â€? negative feedback canâ€™t be present?Â Absurd.

BA:Â Again, this is a major error in your thinking.Â If there were â€œno resistance at all,â€? the force acting against the spring would simply continue to accelerate the mass so long as the force continued.Â Resistance to disturbance is that which opposes the disturbance.Â The reaction force of the spring opposes the disturbance.Â This reaction force is given by Hookâ€™s Law.Â You canâ€™t make this reaction force go away by saying that itâ€™s â€œjust physics.â€?Â The reaction force opposes changes in position induced by the disturbance and by definition thatâ€™s negative feedback.Â Why canâ€™t you understand that?

BA: If there were no â€œnegative feedback, the impressed force would continue to accelerate the mass so long as the force persisted,

RM: Actually, this would only happen if the spring and damper were orbiting the earth (instead of being attached to a solid wall) and the force were applied by thruster engines.

Well yes, of course.Â Your example assumes that the reaction force has been eliminated because the spring is no longer being compressed.Â How does that example â€œproveâ€? that there is no negative feedback in the normal case?Â Answer: It doesnâ€™t.Â In the normal case in which one end of the spring is anchored, the impressed force would also continue to accelerate the mass if there were no reaction force present to oppose that acceleration.Â Youâ€™d have a spring constant of zero, which is equivalent to removing the spring.

but as you can see, this is not what happens. The system develops some degree of resistance to being disturbed from its resting state – enough resistance to bring the mass to a hallt at a new position.

RM: Where do you see resistance to disturbance? As far as I can see, the simulation shows that the disturbance (Force) is completely effective.

BA:Â You reach that conclusion only by defining â€œresistance to disturbanceâ€? in an idiosyncratic way that disagrees with the standard meaning of the phrase.Â Itâ€™s Humpty-Dumpty reasoning – yyou just redefine your terms to make them mean what you need them to mean in order to reach the conclusion you ardently believe in.

BA: At the seven second mark the impressed force suddenly disappears, leaving the counterforces (â€œnegative feedbackâ€?) to accelerate the mass toward the initial resting point. After some oscillation, the damping forces (which are proportional to velocity) convert the remaining energy of the system to heat and the system returns to its initial resting position.

RM: Again, that is exactly the expected effect on the position (Pos) of the mass if the disturbance Force goes to 0.

BA: So here we have a system that resists disturbances (but does not resist them as strongly as a control system does) and returns to its initial equilibrium position once the disturbance is removed.

RM: I see a beautiful demonstration of the fact that a mass-spring system does not put up any resistance to the effect of a disturbance (Force) on a variable (Pos) affected by that disturbance.

BA:Â You see what you want to see, but it isnâ€™t what is happening in the demo.Â The mass-spring-damper system puts up PLENTY of resistance to the effect of the disturbance.Â Somehow, because that resistive force is computed via laws of physics, you think that they donâ€™t exist, or at least donâ€™t count as negative feedback.

BA:Â Oh, and by the way, Iâ€™ve asked you on at least two occasions whether you agree with my definition of â€œnegative feedback.â€? Â Thus far I havenâ€™t gotten a reply.Â I suspect itâ€™s because admitting that the definition is correct destroys your assertion that equilibrium systems do not employ negative feedback – beecause negative feedback is there by definition.

BA:Â At this point Iâ€™ve done all I can do to persuade you of the truth about equilibrium systems.Â If youâ€™re still not convinced, you might want to stop by the physics department next time youâ€™re on campus and discuss the issue with a real physicist. Â It might save you some future embarrassment.

Bruce

[Martin Taylor 2-15.02.06.09.41]

[From Rick Marken (2015.02.05.2300)]

[No quote]

The logic of the argument so far, as I understand it:

RM: All systems with negative feedback are control systems.

BA (and later MT): That's not true, and here are several counter-examples, including equilibrium systems and other examples.

RM: Equilibrium systems don't control, so they don't have negative feedback. (Explicitly saying that he will consider the other examples later, but from here on they are ignored totally).

BA and MT: Here's how to see that the equilibrium systems (pendulum and mass-spring) do exhibit negative feedback.

RM: Those systems don't control, so they don't have negative feedback. Bill Powers proved that all systems with negative feedback are control systems.

MT: Bill Powers did not consider any of the counter-examples, and in any case he made no such claim in the paper cited.

BA and MT: Here are other ways to see that they have negative feedback,

RM: There's no equal and opposite force opposing the disturbance in a pendulum or a spring, it's friction that moves the pendulum and the spring-mass back to their rest position.

MT: Friction opposes movement, at least in normal physics.

RM: Nevertheless I have proved that it is friction that moves the pendulum back to its vertical resting position.

MT: Friction is the way the energy supplied by the disturbance is dissipated to the environment, not force returning the pendulum to vertical.

RM: What's all this nonsense about energy?

RM: These systems don't control, so they don't have negative feedback.

Which is where we are now. RM has the dogmatic position that if a system has negative feedback, it must be a control system, so any system that doesn't control therefore doesn't have negative feedback. No physical or logical demonstration to the contrary can prevail against such blind faith, so there's no point in continuing the discussion.

Martin

[From Bruce Abbott(2014.02.06.1140 EST)]

This is a brief addendum to my previous reply.

Rick Marken (2015.02.05.2300) –

Bruce Abbott (2015.02.05.2120 EST)

RM: The expected effect of the disturbance Force is to move the mass to a position that is proportional to the reciprocal of the spring constant times the Force. That is, the expected value of Pos after the application of the disturbing Force is 1/s(Force). And that seems to be exactly what’s shown in your simulation This is easiest to see if you set the damping to 10 to cut down the transient overshoot and oscillation that is seen when the disturbance Force is applied (and removed) abruptly. When you run the simulation with the spring constant s = 100 Pos ends up about 5 times higher on the Position graph than it does when the spring constant s= 500. Of course, when the disturbance Force goes to 0 the mass returns to the 0 position regardless of the spring constant since 1/s * 0 is zero no matter what the value of s.

BA: This lays to rest any idea that the effect of the â€œnegative feedbackâ€? is evident only after the external force is removed.

RM: Actually, the simulation shows that negative feedback isn’t evident at all. If there were any negative feedback – any resistance at all to the effect of the disturbance Force – then the disturbance force (Force) would have less effect on the position of the mass than expected from physics. But the disturbance Force has exactly the expected effect indicating no negative feedback at all; the mass-spring system is a Z-system (open-loop causal system).

RM: Where do you see resistance to disturbance? As far as I can see, the simulation shows that the disturbance (Force) is completely effective.

RM: I see a beautiful demonstration of the fact that a mass-spring system does not put up any resistance to the effect of a disturbance (Force) on a variable (Pos) affected by that disturbance.

BA: It finally dawned on me why you believe that the mass-spring-damper system shows no resistance to disturbance.Â As I noted previously, your reasoning is faulty, but let me be more explicit about where I believe the fault lies.

BA: You believe that the mass-spring-damper system is a lineal causal system, and as such does not involve negative feedback.Â So when the system comes to rest under continuous applied force, the stable position of the mass is just what the physical computations based on Acc = (Force – s*Pos)/Mass predicts for the level of applied force…Â You then compare the observed deflection of the mass under this force to that predicted by the physics to the deflection predicted by the physical model.Â There is no difference, so you conclude that there is no resistance to deflection and that, therefore, this passive equilibrium system does not employ negative feedback.

BA: In fact the portion of the system equation –s*Pos gives the opposing force, which by definition is negatiive feedback operating against the change in the position of the mass.Â So what we have is a resting position expected from such a negative feedback system.Â What is expected from an open-loop, lineal causal system is not the stable resting position that is observed, itâ€™s runaway acceleration of the mass.Â Comparing the actual behavior to that predicted by a lineal (no feedback) causal system, it is easy to see the dramatic difference between the deflection of the mass expected under the assumption of lineal causality and the behavior this equilibrium system actually demonstrates.

BA:Â Your â€œtest for resistance to disturbanceâ€? reminds me of Monty Pythonâ€™s â€œtest for a witchâ€? in Search for the Holy Grail. Â It goes like this: Witches burn.Â What else burns?Â Wood!Â And what does wood do when you put it in water?Â It floats! And what else floats?Â A duck! Therefore, if a woman weighs as much as a duck . . . .sheâ€™s a witch!Â Your test makes just as much sense.

Bruce

[From Rick Marken (2015.02.06.1200)] ···

Bruce Abbott (2015.02.06.0900 EST)

RM: Yes, the variable called Force is the disturbance to the position, Pos, of the mass. What is clear from the graph is that this **disturbance is completely effective. The disturbance Force moves the mass to the position expected it would be expected to move to if there were no negative feedback at all.Â **

Â

BA: For you, âcompletely ineffectiveâ? means that the system comes to rest (under continued applied force) exactly where one would expect it to come to rest based on the laws of physics.Â

RM: Actually, it means that the disturbance has the expected effect on the variable that is presumably being “stabilized” by the system. In the mass spring system, the position of the mass – called “Pos” in your simulation – is analogous to the controlled variable, q.i, in a control system. In a control system the state of q.i is a function of disturbance variables and the output of the system. In the simplest, linear case we say:

q.i = k.dd - k.oo Â Â Â Â Â Â Â Â Â (1)

where d is the disturbance variable, o is the output of the system and k.d and k.o are the (linear) disturbance and output functions, respectively. The effect of output on the controlled variable, k.oo, is the feedback effect of the control system’s output on its input. If the feedback is negative (such that k.o is negative,as shown in equation 1) Â then the output of the system “compensates” for the effect of the disturbance on q.i so that k.dd will have less effect on q.i than if there were no negative feedback.Â

RM: The equation for the “stabilized” variable in a mass spring system that is analogous to equation 1 is:

Pos = k.dForce - k.oo Â Â Â Â Â (2)

RM: The position, Pos, of the mass is a result of the applied Force (k.dForce) and (you claim) the negative feedback output of the system (k.oo). We know from Hooke’s law that k.d must equal 1/s, which is the spring constant. So without negative feedback (or any feedback) the expected Pos of the mass after applying a known Force to the mass is

Pos = 1/s*Force Â Â Â Â Â Â Â Â Â Â Â (3)

RM: Since this is exactly the result we get with a mass spring system, as demonstrated by your simulation, I conclude that the feedback term, k.o*o, is not necessary to explain the behavior of Pos. There is no need to imagine the existence of negative feedback in a mass-spring system.Â

RM: This analysis is similar to the “quasi-static” analysis of open and closed loop systems in Powers 1978 Psych review paper “Quantitative Analysis of Purposive Systems”, which is reprinted in LCS I. Quasi-static means that the calculations are done with algebraic rather than differential equations. So the analysis leaves out the transient effects of dynamic changes in the variables over time. It is in these transient effects that equilibrium theorists seem to see evidence of negative feedback. But the quasi-static analysis shows that there really is no negative feedback in a mass spring system – in the sense of system output compensating for the effects of disturbance to a variable (Pos in this case).Â

RM: The fact that equation 3 is sufficient to account for the behavior of the mass (variation in its position, Pos) without any need for a negative feedback term (as in equation 2) shows that there is actually no negative feedback involved. Of course, to do a correct dynamic analysis of the behavior of Pos you have to take into account not only the differential equation version of Hooke’s law but also damping (as in your simulation). That let’s you correctly account for the transient dynamic effects of the disturbance to the mass’s position. But equation 3 shows that those effects don’t depend on the existence of negative feedback.There is no negative feedback (in the control theory sense) in the mass-spring system.Â

BA: If you apply this definition [of negative feedback–RM] to the operation of a control system that applies force to a CV in order to control position, then by the same token you must conclude that the control system also comes to rest exactly where one expects it to based on the laws of physics.Â

RM: I disagree. See my explanation above. A control system applies a force (k.oo) to a CV (q.i) that counters the force (k.dd) that is also being applied to the CV by the disturbance. An equilibrium system doesn’t do that with its “stabilized” variable.Â

Â

BA: That is the point at which the counterforce supplied by the control system equals the disturbing force acting on the controlled variable.Â

RM: This suggests that you are thinking of the Force applied to the mass as analogous to the reference specification, r, of control system. I think this must be the same way equilibrium theorists think of control systems as well. This is the view of control systems that we get from motor control theorists; they get the control theory right, they just map it to behavior incorrectly. The main mapping problem is putting the reference specification for the controlled variable outside rather than inside the control system.

RM: I think this is the heart of the problem we have. You are looking at the Force applied to the mass in a mass spring system as analogous to the reference signal in a control system. Force specifies a particular position for the mass – a particular value of Pos. When Force is 0, the reference position of the mass is the resting position; when Force is > 0 the reference position is to the left of the resting position (in your simulation); when Force is <0 the reference position is to the right.Â

RM: When you apply a Force you see that as moving the mass to a new reference position – a new “equilibrium point”. Due to the springiness of the spring the mass “overshoots” the new “reference” or “equilibrium” position immediately after a new Force value is applied. The restoring force of the spring brings the mass back from this overshoot toward the new “reference” or “equilibrium” position. So you see the restoring force of the spring as negative feedback that brings the mass back to the new reference position that was “set” by application of the Force.Â

RM: If you look at a mass spring system this way – with the disturbance Force essentially serving as a reference signal – Â then I can see why you would think of the restoring force as being like negative feedback – an “output” of the spring that compensates for the overshoot that occurs when the reference changes. But this is just grafting control system terminology onto an S-R process. The Force applied to the mass is just a disturbance (stimulus, S) to the position (Pos) of the mass (the response, R) and it’s basic physics that explains the oscillatory approach of Pos to its post disturbance steady state value.

RM: The fact that the Force exerted on the mass in a mass-spring system is not a reference specification for the position (Pos) of the mass can be seen by changing the spring constant in your simulation of the mass spring system. When you do this you will see that the same Force results in quite different values of Pos. So the same “reference” for the state of a variable (mass position, “Pos” in this case) results in very different values of that variable. This would not happen in a negative feedback system, which would compensate for the change in spring constant by appropriately adjusting it’s output, Â bringing the controlled variable to the reference value. So clearly Force is not the reference input to a control system.Â

RM: But now I think I understand why you see equilibrium systems as negative feedback systems. And I also think I now have a better understanding of why equilibrium theorists don’t like PCT. PCT puts reference specifications for "stabilized " (controlled) variables inside the organismÂ – specifically inside the organism’s nervous system. But this implies that organisms have internal purposes. So equilibrium theorists – who are apparently strongly committed to the causal model of behavior-- see in “equilibrium systems” a way to put these specifications in the world outside the organism’s nervous system so that their S-R view of behavior can be preserved.Â So you’ve helped me understand not only what equilibrium theorists are up to but why their ideas are such an enormous disturbance to my concept of how organisms work.Â

RM: The rest of your reply is based on a misunderstanding of the role of Force in your lovely simulation of a mass-spring system so I’ll stop here and wait for further attempts by you to show that equilibrium systems are, indeed, negative feedback systems.

BestÂ

Rick

Â

BA: Just as in the spring-mass-damper system, the counterforce supplied by the control system will limit how much change occurs to position.Â The only difference between the control system and the equilibrium system in this case is in the amount of change in position that the applied force will produce.

Â BA: The equilibrium system being subjected to a continued disturbing force will see the mass move to the point at which the reaction force (as described by Hookâs Law) equals the disturbing force.Â The spring must be compressed a certain distance in order to produce this equal reaction force, and that is the position at which the system at rest will maintain.

Â

BA: Assume a proportional control system for positioning a mass (no springs here).Â This control system, being subjected to a continued disturbing force, will see the mass move to the point at which the counterforce produced by the control system exactly equals the applied force.Â To generate this counterforce the disturbing force must move the mass far enough from its reference position that the error between reference and position, acting on the output function, produces a force equal to the disturbing force.Â The amount of distance required to achieve this equilibrium of forces will depend on the loop gain of the control system.

Â

BA:Â from this comparison it can be seen that the proper definition of âineffective in resisting disturbanceâ? in these two systems is NOT that the position of the mass moves exactly as one would expect given the applicable laws of physics (because the equilibrium system and the control system both do that).Â The proper definition of ineffective would be that the disturbance is unopposed by the system.Â So, the argument you are presenting here is based on adopting a faulty definition of âineffective.â?Â When you adopt the correct definition, it can be seen that both systems are âeffective in resisting the disturbanceâ? but to different degrees.Â As Iâve pointed out ad nauseum, the difference is due to the fact that the equilibrium system gets its restorative force from the energy supplied by the disturbance itself, whereas the control system has an independent source of energy to power its opposing force, and can generate a far stronger opposition to the extent that its loop gain well exceeds 1.0.

Â

RM: The expected effect of the disturbance Force is to move the mass to a position that is proportional to the reciprocal of the spring constant times the Force. That is, the expected value of Pos after the application of the disturbing Force is 1/s(Force). And that seems to be exactly what’s shown in your simulation This is easiest to see if you set the damping to 10 to cut down the transient overshoot and oscillation that is seen when the disturbance Force is applied (and removed) abruptly. When you run the simulation with the spring constant s = 100 Pos ends up about 5 times higher on the Position graph than it does when the spring constant s= 500. Of course, when the disturbance Force goes to 0 the mass returns to the 0 position regardless of the spring constant since 1/s * 0 is zero no matter what the value of s.

Â

BA:Â Yes, itâs exactly what you expect in this system owing to the reaction force that develops as the spring is compressed.Â The reaction force, acting against the effect of disturbing force increases until it equals the disturbing force.Â Thereâs a name for this kind of influence.Â Itâs called ânegative feedback.â?

Â

RM: It would be nice if you could put a line one the chart indicating the expected steady state position of the mass after the application of a non-zero disturbance Force. I think you will find that the position of the mass will end up right on that line after a brief period of damped oscillation.Â

Â

BA: This lays to rest any idea that the effect of the ânegative feedbackâ? is evident only after the external force is removed.Â

Â

RM: Actually, the simulation shows that negative feedback isn’t evident at all. If there were any negative feedback – any resistance at all to the effect of the disturbance Force – then the disturbance force (Force) would have less effect on the position of the mass than expected from physics. But the disturbance Force has exactly the expected effect indicating no negative feedback at all; the mass-spring system is a Z-system (open-loop causal system).Â

Â

BA:Â So if itâs âexpected from physics,â? negative feedback canât be present?Â Absurd.

Â

BA:Â Again, this is a major error in your thinking.Â If there were âno resistance at all,â? the force acting against the spring would simply continue to accelerate the mass so long as the force continued.Â Resistance to disturbance is that which opposes the disturbance.Â The reaction force of the spring opposes the disturbance.Â This reaction force is given by Hookâs Law.Â You canât make this reaction force go away by saying that itâs âjust physics.â?Â The reaction force opposes changes in position induced by the disturbance and by definition thatâs negative feedback.Â Why canât you understand that?

BA: If there were no ânegative feedback, the impressed force would continue to accelerate the mass so long as the force persisted,

RM: Actually, this would only happen if the spring and damper were orbiting the earth (instead of being attached to a solid wall) and the force were applied by thruster engines.Â

Â

Well yes, of course.Â Your example assumes that the reaction force has been eliminated because the spring is no longer being compressed.Â How does that example âproveâ? that there is no negative feedback in the normal case?Â Answer: It doesnât.Â In the normal case in which one end of the spring is anchored, the impressed force would also continue to accelerate the mass if there were no reaction force present to oppose that acceleration.Â Youâd have a spring constant of zero, which is equivalent to removing the spring.

but as you can see, this is not what happens.Â The system develops some degree of resistance to being disturbed from its resting state  enough resistance to bring the mass to a hhalt at a new position.

RM: Where do you see resistance to disturbance? As far as I can see, the simulation shows that the disturbance (Force) is completely effective.Â

Â

BA:Â You reach that conclusion only by defining âresistance to disturbanceâ? in an idiosyncratic way that disagrees with the standard meaning of the phrase.Â Itâs Humpty-Dumpty reasoning  you just redefine yoour terms to make them mean what you need them to mean in order to reach the conclusion you ardently believe in.

BA: At the seven second mark the impressed force suddenly disappears, leaving the counterforces (ânegative feedbackâ?) to accelerate the mass toward the initial resting point.Â After some oscillation, the damping forces (which are proportional to velocity) convert the remaining energy of the system to heat and the system returns to its initial resting position.

RM: Again, that is exactly the expected effect on the position (Pos) of the mass if the disturbance Force goes to 0. Â

Â

Â BA: So here we have a system that resists disturbances (but does not resist them as strongly as a control system does) and returns to its initial equilibrium position once the disturbance is removed.

RM: I see a beautiful demonstration of the fact that a mass-spring system does not put up any resistance to the effect of a disturbance (Force) on a variable (Pos) affected by that disturbance. Â

Â

BA:Â You see what you want to see, but it isnât what is happening in the demo.Â The mass-spring-damper system puts up PLENTY of resistance to the effect of the disturbance.Â Somehow, because that resistive force is computed via laws of physics, you think that they donât exist, or at least donât count as negative feedback.

Â

BA:Â Oh, and by the way, Iâve asked you on at least two occasions whether you agree with my definition of ânegative feedback.â? Â Thus far I havenât gotten a reply.Â I suspect itâs because admitting that the definition is correct destroys your assertion that equilibrium systems do not employ negative feedback  because negativee feedback is there by definition.

Â

BA:Â At this point Iâve done all I can do to persuade you of the truth about equilibrium systems.Â If youâre still not convinced, you might want to stop by the physics department next time youâre on campus and discuss the issue with a real physicist.Â It might save you some future embarrassment.

Â

Bruce

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

[Martin Taylor 2015.02.06.15.29]

``````Perhaps you might think a little about the background to Hooke's
``````

law. Hooke’s law deals with the negative feedback force tht opposes
the applied force. When the two are equal, the mass is in its stable
position.
Not calculable, because without the negative feedback (Hooke’s law)
force, the mass would accelerate to infinity under a continuing
applied force.
But I forgot. You have said that physical law doesn’t apply to a
control system, and only control systems have negative feedback, so
the Hooke’s law force THEREFORE cannot represent negative feedback,
which appears only in systems to which physical laws do not apply.
Wilder and wilder!
Martin

···

[From Rick Marken (2015.02.06.1200)]

``````                Bruce
``````

Abbott (2015.02.06.0900 EST)

``````                      RM: Yes, the variable
``````

called Force is the disturbance to the
position, Pos, of the mass. What is clear from
the graph is that this ** disturbance is
completely effective. The disturbance Force
moves the mass to the position expected it
would be expected to move to if there were
no negative feedback at all.Â **

Â

``````                      BA:
``````

For you, â€œcompletely ineffectiveâ€? means that
the system comes to rest (under continued
applied force) exactly where one would expect
it to come to rest based on the laws of
physics.Â

``````          RM: Actually, it means that the disturbance has the
``````

expected effect on the variable that is presumably being
“stabilized” by the system. In the mass spring system, the
position of the mass – called “Pos” in your simulation –
is analogous to the controlled variable, q.i, in a control
system. In a control system the state of q.i is a function
of disturbance variables and the output of the system. In
the simplest, linear case we say:

q.i = k.dd - k.oo Â Â Â Â Â Â Â Â Â (1)

``````          where d is the disturbance variable, o is the output of
``````

the system and k.d and k.o are the (linear) disturbance
and output functions, respectively. The effect of output
on the controlled variable, k.oo, is the feedback effect
of the control system’s output on its input. If the
feedback is negative (such that k.o is negative,as shown
in equation 1) Â then the output of the system
“compensates” for the effect of the disturbance on q.i so
that k.d
d will have less effect on q.i than if there were
no negative feedback.Â

``````          RM: The equation for the "stabilized" variable in a
``````

mass spring system that is analogous to equation 1 is:

Pos = k.dForce - k.oo Â Â Â Â Â (2)

``````          RM: The position, Pos, of the mass is a result of the
``````

applied Force (k.dForce) and (you claim) the negative
feedback output of the system (k.o
o). We know from
Hooke’s law that k.d must equal 1/s, which is the spring
constant.

``````          So without negative feedback (or any feedback) the
``````

expected Pos of the mass after applying a known Force to
the mass is

[From Rick Marken (2015.02.06.1700)]

···

Bruce Abbott(2014.02.06.1140 EST)–

Â

Â BA: It finally dawned on me why you believe that the mass-spring-damper system shows no resistance to disturbance. Â

Â

BA: You believe that the mass-spring-damper system is a lineal causal system, and as such does not involve negative feedback.Â So when the system comes to rest under continuous applied force, the stable position of the mass is just what the physical computations based on Acc = (Force – s*Pos)/Mass predicts for the level of applieed force.Â

RM: You are absolutely right and I am wrong. I finally see it now thanks to the above equation. Suddenly I realized that Force and Â s*Pos in that equation correspond exactly to the reference signal and perceptual variable in the system equation for a control loop. Here’s the system equation for a control loop:

o = k.o (r-p)

and here is your equation

Acc = k.o (Force - s*Pos)

where k.o = 1/Mass

And the equation that gives position, Pos, as a function of acceleration, Acc, closes the loop.Â

Pos = Pos +(Vel+.5Accdt)*dt

RM: So this certainly can be considered a negative feedback system. The controlled variable is a force – sPos – not the position – Pos – and the reference for this force is provided by the externally applied force, Force. And it does behave like a negative feedback system because it compensates for the only disturbance that can enter this loop: variations in the spring constant, s. I believe this is shown in your simulation. When the spring constant is changed, Pos is changed also so that sPos comes to match the reference, Force. It would be nice to make s variable and show a plot of s, Acc and sPos over time while Force is constant; you should see something like the “mirror image” plots that we get in a tracking task: with s in the role of d, Acc in the role of o and sPos in the role of q.i.Â

RM: So, my gosh, it seems that you and Martin are right; a mass spring system seems to be a negative feedback system. Moreover, it seems to be a control system; the controlled variable is force (not position) and the reference for force is set by the external force, Force, applied to the mass. If you, Bruce, could change your simulation to show this I think it would be really cool. That is, add the ability to vary the frequency and amplitude of variations in s throughout a trial (like the one you have now where there is just one change in the reference Force); you already plot Acc so I would just have the plot show Acc, s and s*Pos (scaled appropriately for display) on the same graph.Â

BestÂ

Rick

You then compare the observed deflection of the mass under this force to that predicted by the physics to the deflection predicted by the physical model.Â There is no difference, so you conclude that there is no resistance to deflection and that, therefore, this passive equilibrium system does not employ negative feedback.

Â

BA: In fact the portion of the system equation –s*Pos gives the opposing force, which by definition is negatiive feedback operating against the change in the position of the mass.Â So what we have is a resting position expected from such a negative feedback system.Â What is expected from an open-loop, lineal causal system is not the stable resting position that is observed, itâ€™s runaway acceleration of the mass.Â Comparing the actual behavior to that predicted by a lineal (no feedback) causal system, it is easy to see the dramatic difference between the deflection of the mass expected under the assumption of lineal causality and the behavior this equilibrium system actually demonstrates.

Â

BA:Â Your â€œtest for resistance to disturbanceâ€? reminds me of Monty Pythonâ€™s â€œtest for a witchâ€? in Search for the Holy Grail.Â It goes like this: Witches burn.Â What else burns?Â Wood!Â And what does wood do when you put it in water?Â It floats! And what else floats?Â A duck! Therefore, if a woman weighs as much as a duck . . . .sheâ€™s a witch!Â Your test makes just as much sense.

Â

Bruce

Â

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

[From Bruce Abbott (2015.02.06.1100 EST)]

Rick Marken (2015.02.06.1700)]

Bruce Abbott(2014.02.06.1140 EST)–

BA: It finally dawned on me why you believe that the mass-spring-damper system shows no resistance to disturbance.

BA: You believe that the mass-spring-damper system is a lineal causal system, and as such does not involve negative feedback. So when the system comes to rest under continuous applied force, the stable position of the mass is just what the physical computations based on Acc = (Force – s*Pos)/Mass predicts for the level of applied force. <

RM: You are absolutely right and I am wrong. I finally see it now thanks to the above equation. Suddenly I realized that Force and s*Pos in that equation correspond exactly to the reference signal and perceptual variable in the system equation for a control loop. Here’s the system equation for a control loop:

o = k.o (r-p)

and here is your equation

Acc = k.o (Force - s*Pos)

where k.o = 1/Mass

And the equation that gives position, Pos, as a function of acceleration, Acc, closes the loop.

Pos = Pos +(Vel+.5Accdt)*dt

Pure nonsense.Â Hereâ€™s the correct system diagram: For clarity Iâ€™m assuming that the system has reached a steady state under the applied force. Â Velocity is therefore zero and the retarding force produced by the damper, which is proportional to velocity, is therefore zero.Â This steady-state condition allows me to remove the velocity feedback, thus simplifying the diagram.

As you can see from the diagram, the applied force is not the reference signal (as you claim it to be), it acts to disturb the position of the mass.Â What corresponds to the reference is the resting position of the mass when no disturbing force is acting on the spring.Â The difference between this position and the current position is given here as â€œeâ€?; it represents the amount of compression of the spring.Â The output function multiplies this displacement by s, the spring coefficient.Â The output is a force proportional to the displacement of the spring.Â This output force of the spring acts to oppose the effect of the applied force on the acceleration of the mass, thus the feedback function simply multiplies the output force by -1.

On the left-hand side of the diagram, the acceleration of the mass produced by the net force acting on the mass (Applied force â€“ feedback force) is integrated in the first box to yield the velocity of the mass; velocity is then integrated in the second box to yield the position of the mass.Â The position of the mass would be called the controlled perception if this were a control system.Â In fact the diagram appears to be identical to a control-system diagram.Â However, there are important differences.Â The energy stored in the spring by its compression provides the output force that opposes spring compression.Â Because of that, the restorative force cannot exceed the disturbing force, which provides the energy to compress the spring.Â This limits system gain to no greater than 1.0 (in the ideal frictionless case) and less than one in any real system where energy is lost due to friction.Â In contrast, the control system draws its output force from an independent source and therefore can potentially be very high, not limited by the energy in the disturbance.Â The â€œreferenceâ€? of the mass-spring-damper is simply the resting position of the mass when no force is being applied to it.Â Its value is determined by the length of the (uncompressed) spring.Â In a control system the position reference can be set at will by the user, even to a value in which the spring is compressed or stretched.

Except for the deletion of force feedback from the damper, the above diagram is the same as the Simulink diagram I found on the internet and posted here a few days ago.Â Iâ€™m not making this stuff up.

Oh, and Iâ€™m STILL waiting for your definition of negative feedback.Â Are we to have it anytime soon?

Bruce

[Martin Taylor 2015.02.06.23.08]

``````Well, at least in that we all agree.
``````

Yes.
No. The energy supplied to the spring comes only from the
disturbance, and is stored in the spring extension or compression,
or in the case of the pendulum is stored as gravitational potential
energy, ready to be released when the disturbance is removed. In a
control system, the energy comes from a separate source. In an
equilibrium system, the system comes to its lowest-energy
configuration. In a control system the controlled variable comes to
a reference value.
Martin
Martin

···

[From Rick Marken (2015.02.06.1700)]

``````                Bruce
``````

Abbott(2014.02.06.1140 EST)–

``````                   BA: It
``````

finally dawned on me why you believe that the
mass-spring-damper system shows no resistance to
disturbance.

``````                  BA:
``````

You believe that the mass-spring-damper system is
a lineal causal system, and as such does not
involve negative feedback. So when the system
comes to rest under continuous applied force, the
stable position of the mass is just what the
physical computations based on Acc = (Force –
s*Pos)/Mass predicts for the level of applied
force.

RM: You are absolutely right and I am wrong.

``````          RM: So, my gosh, it seems that you and Martin are
``````

right; a mass spring system seems to be a negative
feedback system.

Moreover, it seems to be a control system

[philip.2.7.10:39am]

let me try a stab at this.

Bill Powers proved that all systems with negative feedback are control systems.

so, feedback means that the value of a variable at a certain point in simulated time changes because of the causal effects propagating in a loop. negative feedback means the variable becomes smaller and pos means that it becomes larger.

a spring with a constant applied force generates a positive feedback situation, because more compression leads to more force and the force is constantly compressing the spring.

therefore,

I think this is an example involving positive feedback “control” and not negative feedback “control”. I think analyzing the final position of the mass is not as important as analyzing the differential changes which are occuring at each moment. I think the acceleration is under negative feedback at all times while the force is under positive feedback at all times. I think the unit of resistance to disturbance of position (caused by force) is negative acceleration. and I think the final position of the spring is simply a side effect of the fact that a force balance is reached and acceleration becomes zero.

furthermore,

I don’t think a spring is a control system at any level higher than the individual atoms which compose the spring are control systems. I think a spring is a highly artificial device not normally found in nature. and I think that it needs to be taken into account that since humans create the spring, they can design it to do whatever they want. I think Rick wants the spring to return back to position zero.

[Martin Taylor 2015.02.07.14.00]

``````That's a myth. Bill never claimed to have proved that, so far as I
``````

know. What Bill did say is that if the value of something is sensed
in one place and influenced in another, and if the sensed value
affects the influenced value in such a way as to reduce the effect
of changes in the sensed value, then the loop is a control system.
The direction of the self-influence of the changes in the sensed
value defines the loop as having negative feedback, and the fact
that this reduces the influence of external factors on the value
sensed defines it as a control system. Therefore negative feedback
systems that fit this description are control systems.
That’s not the same as saying all negative feedback systems are
control systems.
To make the equivalent statement about a control system, the more
force the disturbance applies to the thing sensed, the more force
the output applies to the thing sensed, so the control system is a
positive feedback system.
I don’t think so.
Martin

···

[philip.2.7.10:39am]

let me try a stab at this.

``````      Bill
``````

Powers proved that all systems with negative feedback are
control systems.

``````        so, feedback means that the value of a
``````

variable at a certain point in simulated time changes
because of the causal effects propagating in a loop.
negative feedback means the variable becomes smaller and pos
means that it becomes larger.

``````        a spring with a constant applied force
``````

generates a positive feedback situation, because
more compression leads to more force and the force is
constantly compressing the spring.

[philip2.7.11:18]

To make the equivalent statement about a control system, the more force the disturbance applies to the thing sensed, the more force the output applies to the thing sensed, so the control system is a positive feedback system.

but I don’t think the spring is a position-control system. because the individual atoms cannot be held responsible for controlling the final position of the spring. it’s just too much responsibility. each atom can only be seen to act with its neighbors.

[philip.2.7.11:40]

the control system is a positive feedback system.

two control systems in conflict constitute a positive feedback situation

[From Rick Marken (2015.02.07.1400)]   ···

:

Bruce Abbott (2015.02.06.1100 EST)–

RM: You are absolutely right and I am wrong. I finally see it now thanks to the above equation. Suddenly I realized that Force and s*Pos in that equation correspond exactly to the reference signal and perceptual variable in the system equation for a control loop. Here’s the system equation for a control loop:

o = k.o (r-p)

and here is your equation

Acc = k.o (Force - s*Pos)

where k.o = 1/Mass

And the equation that gives position, Pos, as a function of acceleration, Acc, closes the loop.

Pos = Pos +(Vel+.5Accdt)*dt

BA: Pure nonsense.

RM: Well, it turns out that you are kind of right on that too. While these equations are superficially similar to the simultaneous equations that define a negative feedback loop, they don’t function as a negative feedback loop. In particular, they don’t lead to control of the apparent controlled variable, s*Pos, nor do they compensate for what I thought might be a disturbance variable, s.

RM: If you include the damping factor in the system equation, however, so that Acc = k.o (Force -dVel- sPos) then these equations do function like a negative feedback loop inasmuch as they bring the apparent controlled variable, s*Pos, to the reference state specified by Force. But, of course, this only happens when s is a constant and the damping is adjusted properly. So I was certainly wrong about s functioning like an independent disturbance; it couldn’t possibly be since it is not independent of Pos.

RM: Here’s a plot of the behavior of the system:

RM: Note that when the reference specification (Force) changes to a new value the system brings the input variable, sPos, into a match. So when reference Force goes from 0 to 10 so does sPos; when the reference returns to 0 so does sPos. So sPos acts like a controlled variable. But when you introduce disturbances it doesn’t.

RM: So you could see the mass spring system as a negative feedback system according to the equations above. There is apparent negative feedback because the output of the system (Acc) has an effect on the input (s*Pos) that is negative. That’s my definition of a negative feedback system, by the way, Bruce; it’s a system where the output of the system has a negative (reductive) effect on the cause of that output.

RM: So the equations above show that you guys are, indeed, correct to say that an equilibrium system can be considered a negative feedback system. And it can be seen as a system that functions like a negative feedback system in the sense that it brings a variable (s*Pos in the case of the mass-spring system) to a reference value “set” by an outside force (Force in this case). But unlike a true negative feedback system it does not protect the controlled variable from the effects of disturbance.

BA: Here’s the correct system diagram:

RM: Thanks for this. Assuming this is the correct functional diagram of a mass spring system I implemented it as a working model and found that it doesn’t seem to work like a negative feedback system. Here’s the code I used, based on your diagram:

e = (r - Pos)

OutputForce = s * e

FeedbackForce = -1 * OutputForce

Acc = (Force - FeedbackForce) / Mass

Pos = Pos + (Vel + 0.5 * Acc * dt) * dt

Vel = Vel + Acc * dt

The result looks like this:

RM: The Force value is exaggerated (it actually went from 0 to 10 and back) so you could see it more clearly on the graph. The reference is set to 0 which is the resting position of the mass. Everything is going fine – the mass is just resting – until the force disturbance is introduced. Then the purported controlled variable (Pos) starts oscillating; and it’s not even the constant amplitude oscillation that occurs in a mass spring system without damping. The oscillation gets bigger and bigger, as though there were positive feedback. It even persists when disturbance force is removed. I thought this might be because the sign of the feedback was wrong so I changed ForceFeedback from FeedbackForce = -1 * OutputForce to FeedbackForce = 1 * OutputForce. This just made things worse - exponentially worse.

RM: Perhaps I derived the working model incorrectly from your diagram. If so I’d appreciate it if you could post the correct code. But at the moment it looks to me that my model of the mass spring system as a negative feedback system (with Force as the reference rather than the disturbance) gives a more accurate representation of what actually happens when you push the mass. It moves exactly to a position proportional (by the spring constant) to the applied force.

BA: Except for the deletion of force feedback from the damper, the above diagram is the same as the Simulink diagram I found on the internet and posted here a few days ago. I’m not making this stuff up.

RM: Well, you are either making something up, there is something wrong with your diagram or I have incorrectly implemented it as a computer program because my implementation of the functional diagram doesn’t produce anything like the behavior produced by the Simulink model of the mass spring system.

BA: Oh, and I’m STILL waiting for your definition of negative feedback. Are we to have it anytime soon?

RM: I’ve given it several times before as well as earlier in this post but here it is again: negative feedback exists when the output of a system has a negative effect on the cause of that output. The equations for my version of the mass-spring modle seem to satisfy that definition:

Acc = k.o (Force - s*Pos)

Pos = Pos +(Vel+.5Accdt)*dt

RM: So the output of the system (Acc) has a negative effect on itself via it’s effect on Pos. So the mass spring system is a negative feedback system that doesn’t seem to resist disturbances; that is, it doesn’t control. It behaves more like it reacts to external forces.

Best

Rick

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

[From Bruce Abbott (2015.02.07.1910 EST)]

[From Rick Marken (2015.02.07.1400)]

Bruce Abbott (2015.02.06.1100 EST)—

RM: Well, it turns out that you are kind of right on that too. While these equations are superficially similar to the simultaneous equations that define a negative feedback loop, they don’t function as a negative feedback loop. In particular, they don’t lead to control of the apparent controlled variable, s*Pos, nor do they compensate for what I thought might be a disturbance variable, s.

The variable s*Pos is the restorative force generated in the spring by its compression.Â Itâ€™s part of the â€œoutputâ€? of the system.Â That explains why it doesnâ€™t function as a CV.

RM: If you include the damping factor in the system equation, however, so that Acc = k.o (Force -dVel- sPos) then these equations do function like a negative feedback loop inasmuch as they bring the apparent controlled variable, s*Pos, to the reference state specified by Force. But, of course, this only happens when s is a constant and the damping is adjusted properly. So I was certainly wrong about s functioning like an independent disturbance; it couldn’t possibly be since it is not independent of Pos.

Force acts as the disturbance to the system.Â That explains why it doesnâ€™t function as a reference.Â The equivalent of a reference in this system is the resting position of the mass when Force = 0.0.

RM: Here’s a plot of the behavior of the system: RM: Note that when the reference specification (Force) changes to a new value the system brings the input variable, sPos, into a match. So when reference Force goes from 0 to 10 so does sPos; when the reference returns to 0 so does sPos. So sPos acts like a controlled variable. But when you introduce disturbances it doesn’t.

RM: So you could see the mass spring system as a negative feedback system according to the equations above. There is apparent negative feedback because the output of the system (Acc) has an effect on the input (s*Pos) that is negative. That’s my definition of a negative feedback system, by the way, Bruce; it’s a system where the output of the system has a negative (reductive) effect on the cause of that output.

RM: So the equations above show that you guys are, indeed, correct to say that an equilibrium system can be considered a negative feedback system. And it can be seen as a system that functions like a negative feedback system in the sense that it brings a variable (s*Pos in the case of the mass-spring system) to a reference value “set” by an outside force (Force in this case). But unlike a true negative feedback system it does not protect the controlled variable from the effects of disturbance.

BA: Hereâ€™s the correct system diagram: RM: Thanks for this. Assuming this is the correct functional diagram of a mass spring system I implemented it as a working model and found that it doesn’t seem to work like a negative feedback system. Here’s the code I used, based on your diagram:

e = (r - Pos)

OutputForce = s * e

FeedbackForce = -1 * OutputForce

Acc = (Force - FeedbackForce) / Mass

Pos = Pos + (Vel + 0.5 * Acc * dt) * dt

Vel = Vel + Acc * dt

The result looks like this: RM: The Force value is exaggerated (it actually went from 0 to 10 and back) so you could see it more clearly on the graph. The reference is set to 0 which is the resting position of the mass. Everything is going fine – the mass is just resting – until the force disturbance is introduced. Then the purported controlled variable (Pos) starts oscillating; and it’s not even the constant amplitude oscillation that occurs in a mass spring system without damping. The oscillation gets bigger and bigger, as though there were positive feedback. It even persists when disturbance force is removed. I thought this might be because the sign of the feedback was wrong so I changed ForceFeedback from FeedbackForce = -1 * OutputForce to FeedbackForce = 1 * OutputForce. This just made things worse - exponentially worse.

BA:Â I explain the observed behavior of the system – beffore you changed the sign of the feedback – below.Â After you cchanged it you had positive feedback, so naturally things got worse.

RM: Perhaps I derived the working model incorrectly from your diagram. If so I’d appreciate it if you could post the correct code. But at the moment it looks to me that my model of the mass spring system as a negative feedback system (with Force as the reference rather than the disturbance) gives a more accurate representation of what actually happens when you push the mass. It moves exactly to a position proportional (by the spring constant) to the applied force.

BA: Except for the deletion of force feedback from the damper, the above diagram is the same as the Simulink diagram I found on the internet and posted here a few days ago. Iâ€™m not making this stuff up.

RM: Well, you are either making something up, there is something wrong with your diagram or I have incorrectly implemented it as a computer program because my implementation of the functional diagram doesn’t produce anything like the behavior produced by the Simulink model of the mass spring system.

BA:Â To simplify the diagram I omitted the contribution of damping to the output.Â Without damping the mass-spring system goes into a stable oscillation about the equilibrium value.Â This would be expected in the ideal case of no energy loss from the system due to, e.g., friction.

BA:Â Your simulation, however, depicts a system whose oscillations grow with time, indicating that the systemâ€™s energy is continually increasing.Â This is an impossibility in the real world where the energy powering the oscillations came from the disturbing force.Â The reason it shows up in the digital simulation is that the computer cannot represent the values of the variables accurately to an infinite number of decimals.Â These tiny errors cumulate over iterations due to the integration used to compute velocity from change in acceleration and position from change velocity.Â The problem is especially noticeable because of this double integration.

BA: Â You can fix this problem by capping the output force (s*Pos) to be no greater than the disturbance force as follows:

Â Â Â Â Â Â Â Â Â Â Â SpringForce = s*Pos

If SpringForce > Force then SpringForce = Force

BA: With this fix added the system will still oscillate but the size of the excursions will not grow.Â It makes sure that the system gain cannot exceed 1.0. Â Â

BA: To allow the oscillations to decay away, simply add the velocity-dependent force generated by the damper to the output:

DamperForce := d*Vel

BA:Â Now add the two forces together:

OutputForce = SpringForce + DamperForce

With these changes, the modified code should look like this:

e = (r - Pos)

SpringForce = s * e

If SpringForce > Force then SpringForce = Force

DamperForce = d * Vel

OutputForce = SpringForce + DamperForce

FeedbackForce = -1 * OutputForce

Acc = (Force - FeedbackForce) / Mass

Pos = Pos + (Vel + 0.5 * Acc * dt) * dt

Vel = Vel + Acc * dt

BA:Â By the way, with damping added you can get away with eliminating the cap on SpringForce so long as the damping coefficient is large enough to â€œleak awayâ€? the cumulating computational errors due to the double integration.

BA: Oh, and Iâ€™m STILL waiting for your definition of negative feedback. Are we to have it anytime soon?

RM: I’ve given it several times before as well as earlier in this post but here it is again: negative feedback exists when the output of a system has a negative effect on the cause of that output.

BA:Â O.K., that will work.Â (I would state it more generally: Negative feedback exists when a change in a variable produces an effect that opposes that change.)

The equations for my version of the mass-spring modle seem to satisfy that definition:

Acc = k.o (Force - s*Pos)

Pos = Pos +(Vel+.5Accdt)*dt

RM: So the output of the system (Acc) has a negative effect on itself via its effect on Pos. So the mass spring system is a negative feedback system that doesn’t seem to resist disturbances; that is, it doesn’t control. It behaves more like it reacts to external forces.

BA:Â I would agree with this except for the part about â€œdoesnâ€™t seem to resist disturbances.â€?Â To see whether the system resists disturbances, open the loop (i.e., remove the feedback).Â You will get runaway acceleration of the mass caused by the applied force (disturbance).Â With feedback, the system does resist the disturbance, but acts like a control system with a gain of less than 1.0.

BA:Â Have we come to an agreement?

Bruce

[Martin Taylor 2015.02.07.17.35]

``````Why do you keep reiterating what has been agreed by all from the
``````

get-go, as if it were a critical argument against the fact that
equilibrium systems are (as they indeed are and have been shown to
be) negative feedback systems?
EQUILIBRIUM SYSTEMS DO NOT CONTROL!!! We get it, have got it,
never have had a problem with it. It’s just a fact.
So please stop using it as though it were an argument.
That’s a funny definition. “The cause” of the output of a control
system is the combination of the output function and the history of
the error variable (two causes). Is that not so? My preferred definition deals with changes in variables, not their
causes, nor their values, and says nothing about what variable you
are considering, something like this: The actual words might be better put, but that’s my idea of negative
feedback.
In the case of the spring, changing the value of the variable
“position of the mass” changes the compression or extension of the
spring, which results in an influence that opposes the change of
position. (See attached figure – it doesn’t seem to want to go
in-line).
Martin ···

[From Rick Marken (2015.02.07.1400)]
:So the mass spring system is a
negative feedback system that doesn’t seem to resist
disturbances; that is, it doesn’t control.

** negative feedback exists when the output
of a system has a negative effect on the cause of that output.**

• `````` Negative feedback exists
``````

when a change in a variable result in an influence that opposes
the change.*

[Philip 5.18.15]

Instead of running the TCV on the position of a mass on a spring, why don’t we run this TCV analysis on a gyroscope’s orientation? This would make for a much richer discussion, because the physical laws are more complex. Perhaps we could even devise a controller similar to the one Bill described for controlling the position of a mass. Angular orientation would substitute for mass position, angular velocity for mass velocity, and torque for applied force.

···

On Sun, Feb 8, 2015 at 7:07 AM, Martin Taylor csgnet@lists.illinois.edu wrote:

[Martin Taylor 2015.02.07.17.35]

[From Rick Marken (2015.02.07.1400)]

** negative feedback exists when the output
of a system has a negative effect on the cause of that output.**

``````Why do you keep reiterating what has been agreed by all from the
``````

get-go, as if it were a critical argument against the fact that
equilibrium systems are (as they indeed are and have been shown to
be) negative feedback systems?

``````EQUILIBRIUM SYSTEMS DO NOT CONTROL!!!!!  We get it, have got it,
``````

never have had a problem with it. It’s just a fact.

``````So please stop using it as though it were an argument.

That's a funny definition. "The cause" of the output of a control
``````

system is the combination of the output function and the history of
the error variable (two causes). Is that not so?

``````My preferred definition deals with changes in variables, not their
``````

causes, nor their values, and says nothing about what variable you
are considering, something like this: * Negative feedback exists
when a change in a variable result in an influence that opposes
the change.*

``````The actual words might be better put, but that's my idea of negative
``````

feedback.

``````In the case of the spring, changing the value of the variable
``````

“position of the mass” changes the compression or extension of the
spring, which results in an influence that opposes the change of
position. (See attached figure – it doesn’t seem to want to go
in-line).

``````Martin
``````
``````        :So the mass spring system is a
``````

negative feedback system that doesn’t seem to resist
disturbances; that is, it doesn’t control.

Here’s a conversation we had where we used a simulator to model a disturbance to a mass on a spring and we were discussing what the controlled variable was - I don’t think we reached a conclusion, but the conversation pivoted between Rick and Bruce. What I propose is to run a simulation of the experiment where the mass on the spring is subject to circular motion and it is the centripetal motion of the mass (i.e. a moment of inertia) which causes the linear displacement of the spring.

···

On Mon, May 18, 2015 at 11:20 AM, PHILIP JERAIR YERANOSIAN pyeranos@ucla.edu wrote:

[Philip 5.18.15]

Instead of running the TCV on the position of a mass on a spring, why don’t we run this TCV analysis on a gyroscope’s orientation? This would make for a much richer discussion, because the physical laws are more complex. Perhaps we could even devise a controller similar to the one Bill described for controlling the position of a mass. Angular orientation would substitute for mass position, angular velocity for mass velocity, and torque for applied force.

On Sun, Feb 8, 2015 at 7:07 AM, Martin Taylor csgnet@lists.illinois.edu wrote:

[Martin Taylor 2015.02.07.17.35]

[From Rick Marken (2015.02.07.1400)]

** negative feedback exists when the output
of a system has a negative effect on the cause of that output.**

``````Why do you keep reiterating what has been agreed by all from the
``````

get-go, as if it were a critical argument against the fact that
equilibrium systems are (as they indeed are and have been shown to
be) negative feedback systems?

``````EQUILIBRIUM SYSTEMS DO NOT CONTROL!!!!!  We get it, have got it,
``````

never have had a problem with it. It’s just a fact.

``````So please stop using it as though it were an argument.

That's a funny definition. "The cause" of the output of a control
``````

system is the combination of the output function and the history of
the error variable (two causes). Is that not so?

``````My preferred definition deals with changes in variables, not their
``````

causes, nor their values, and says nothing about what variable you
are considering, something like this: * Negative feedback exists
when a change in a variable result in an influence that opposes
the change.*

``````The actual words might be better put, but that's my idea of negative
``````

feedback.

``````In the case of the spring, changing the value of the variable
``````

“position of the mass” changes the compression or extension of the
spring, which results in an influence that opposes the change of
position. (See attached figure – it doesn’t seem to want to go
in-line).

``````Martin
``````
``````        :So the mass spring system is a
``````

negative feedback system that doesn’t seem to resist
disturbances; that is, it doesn’t control.

[Martin Taylor 2016.08.07.23.21]

``````Given the equivalence between acceleration and gravity, which I
``````

believe has been tested out to quite a few decimal places, what
would be the point? How would it differ from having a heavier weight
on the pendulum? Or were you thinking of introducing angular
accelerations into the mix? If so, what would they tell you?
Martin

···

On 2016/08/7 8:40 PM, PHILIP JERAIR
YERANOSIAN wrote:

``````    Here's a conversation we had where we used a
``````

simulator to model a disturbance to a mass on a spring and we
were discussing what the controlled variable was - I don’t think
we reached a conclusion, but the conversation pivoted between
Rick and Bruce. What I propose is to run a simulation of the
experiment where the mass on the spring is subject to circular
motion and it is the centripetal motion of the mass (i.e. a
moment of inertia) which causes the linear displacement of the
spring.

``````      On Mon, May 18, 2015 at 11:20 AM,
``````

PHILIP JERAIR YERANOSIAN pyeranos@ucla.edu wrote:

[Philip 5.18.15]

``````          Instead of running the TCV on the position of a mass on a
``````

spring, why don’t we run this TCV analysis on a
gyroscope’s orientation? This would make for a much
richer discussion, because the physical laws are more
complex. Perhaps we could even devise a controller
similar to the one Bill described for controlling the
position of a mass. Angular orientation would substitute
for mass position, angular velocity for mass velocity, and
torque for applied force.

``````                On Sun, Feb 8, 2015 at 7:07
``````

AM, Martin Taylor csgnet@lists.illinois.edu
wrote:

``````                    [Martin
``````

Taylor 2015.02.07.17.35]

``````                        [From Rick Marken
``````

(2015.02.07.1400)]
:So the mass
spring system is a negative feedback
system that doesn’t seem to resist
disturbances; that is, it doesn’t
control.

** negative feedback
exists when the output of a system has a
negative effect on the cause of that
output.**

``````                    Why do you keep reiterating what has been agreed
``````

by all from the get-go, as if it were a critical
argument against the fact that equilibrium
systems are (as they indeed are and have been
shown to be) negative feedback systems?

``````                    EQUILIBRIUM SYSTEMS DO NOT CONTROL!!!!!  We get
``````

it, have got it, never have had a problem with
it. It’s just a fact.

``````                    So please stop using it as though it were an
``````

argument.

``````                     That's a funny definition. "The cause"
``````

of the output of a control system is the
combination of the output function and the
history of the error variable (two causes). Is
that not so?

``````                    My preferred definition deals with changes in
``````

variables, not their causes, nor their values,
and says nothing about what variable you are
considering, something like this: * Negative
feedback exists when a change in a variable
result in an influence that opposes the
change.*

``````                    The actual words might be better put, but that's
``````

my idea of negative feedback.

``````                    In the case of the spring, changing the value of
``````

the variable “position of the mass” changes the
compression or extension of the spring, which
results in an influence that opposes the change
of position. (See attached figure – it doesn’t
seem to want to go in-line).

``````                    Martin
``````