# Knot on dot [Was Perceived Present [was Shared references]]

When I control “knot on the dot” and the “knot is
on the dot” I don’t affect the

environment by any actions. The reference is a1, the perceptual
signal is a1, the

error is zero, the feedback signal is zero (I think) and the
disturbance "knot on

the dot" results in a perceptual signal a1. This is
continued without any affects

till the “knot is not on the dot”

.

[Martin]>Huh? There’s no perceptual signal if the knot is on the
dot?

Did I say that?
[From Bill Powers (2004.08.05.0845 MDT)]
Bjorn Simonsen (2004.08.05,12:18
EuST)–

[From Bjorn Simonsen (2004.08.04,09:30 EuST)]
p=
ki*ke*ko*e +kd*d. I meant to
say that when knot is on the dot, e=0. Then p = kd*d.
I say “huh?” too. What happened to the effect of the output on
qi? The knot is in the environment, and qo has to affect it
continuously to counteract the continuing effect of d. If d is a
steady pull on one rubber band, qo has to be a steady pull in the other
direction on the other rubber band. The force produced by the output is
certainly an effect on the environment, isn’t it?
Let’s analyze the rubber band experiment. To make the analysis easy,
let’s assume that the rubber bands are identical, so the knot will always
be halfway between their movable ends. I’ll also ignore the resting
length of the rubber bands, saying it is zero. Notice that we’re defining
all the environmental variables as positions, so their values can be
added algebraically. This means that we have a much simplified dependence
of the knot position, qi, on the disturbance position d and the output
position qo.

1. qi = (qo + d)/2
That’s the environment equation.
The controller equations are
p = qi
qo = Ko*(r - p), which reduces to
2. qo = Ko*(r - qi)
We’ll solve for qo by subsituting the right side of equation 1 for
qi:
qo = Ko*[r - (qo+d)/2]
Expanding:
qo = Kor - Koqo/2 - Kod/2
Collecting terms in qo on the left:
qo + qo
Ko/2 = Kor - Kod/2
Extracting the common term qo on the left:
qo*(1 + Ko/2) = Kor - Kod/2
Dividing both sides by (1 + Ko/2):
Kor - Kod/2
qo = --------------- .
1 + Ko/2
For an exact solution we can subsitute specific values for d and Ko and
see what qo will be. First, however, a value for r is needed. The
reference condition for qi is “knot over dot.” If we place the
dot conveniently at the zero position of our measurement scale, The
desired position of qi is zero, which is where the dot is, so the
reference signal is equal to zero: that makes the solution even
simpler.
Let’s say r = 0, Ko = 100, and d = -20 cm (20 centimeters in the negative
direction)
0 cm * 100 + 20 cm * 100/2
qo = --------------------------
(1 +
100/2)
qo = 1000/51 cm
qo = 19.61 cm
The position of the knot, qi, will be (by equation 1) the average of qo
and d, or
qi = (19.61 - 20)/2 or
qi = -0.195 cm
So this proportional control system with an output gain of 100 can keep
the knot within 2 mm of the center of the dot when the disturbance is 20
cm, and even closer when the disturbance is smaller.
Exercise 1: Redo the calculation with an output gain of 2000. What are
the values of qo and qi?
Exercise 2: Redo the calculation with the reference condition set to
“knot 5 cm to the right of the dot” (r = 5.00).
Exercise 3: How big is the error when Ko becomes infinite?
Exercise 4: How will this result be affected if the rubber bands are
identical but their elastic constants are nonlinear (i.e., change with
the amount of stretch)?
Exercise 5: Derive the control system equations for the case when the
rubber bands have different resting lengths and/or different elastic
constants Kd and Ke. To do this exercise you have to do some physical
analysis of the environment and the equations will look considerably more
complex. The results, however, will be the same if you express the output
as output position minus resting length, and the disturbance as the
disturbance position minus resting length. Assume that the disturbance
position is never less than the resting length of the associated rubber
band.
···

It was this view of a proportional control system that led William Ashby
to say that a compensating system was in principle a better control
system than a negative feedback control system. The compensating system
can be adjusted to make the effect of a disturbance on a controlled
variable zero, whereas a proportional negative feedback control system
could not – not quite.
This shows how limited Ashby’s understanding of control systems was. In
the first place, as exercise 3 shows, the error can be made as small as
desired in a negative feedback control system (if it remains stable)
simply by raising the gain. And second of all, the system will remain
stable if, for any gain, a slowing factor is inserted that reduces the
amount of error correction that takes place per unit time. Such a slowing
factor is needed when there is a transport lag in the system.
Finally, in the limiting case of infinite gain, the required slowing
factor turns the output function into a pure integrator with a suitable
gain factor. And when the output function is a pure integrator, the error
that remains when the system reaches equilibrium will be exactly
zero
– just as for the ideal compensating system.

However, when the error is zero in a system with an integrating output
function, the output is not zero, unlike the situation in a proportional
control system. It is some steady value, just enough to cancel exactly
the effect of a constant disturbance on the input quantity. If a positive
error develops, the output will integrate upward, increasing until the
error is again exactly zero. A negative error will produce the opposite
action, thye output integrating downward until there is again exactly
zero error in the steady state.

Of course when the disturbance changes, it will take a little time for
the output to cancel it exactly once again. And if the disturbance
changes continuously, the output will never quite catch up. But remember
why we need the integrator in the first place: because there is a
transport lag in the system. Without the transport lag, or with a very
very small transport lag, we could crank Ko up toward infinity and get
essentially zero error (there is zero transport lag in the algebraic
equations above: all variables in a system of algebraic equations change
at literally the same time: there is no time). By adjusting the slowing
factor, we can make the error correction as fast as possible for any
given amount of transport lag.

Obviously, a compensating system with a transport lag would also have to
allow error while the disturbance was changing, so it cannot act
perfectly in those circumstances any more than the negative feedback
control system can. If one can postulate a compensating system with no
internal tranport lag, then one can postulate a negative feedback control
system with no internal tranport lag. However you slice it, the result
comes out that the compensating system is no faster than the negative
feedback control system, and keeps the error in the controlled quantity
no smaller, even in principle.

There are many aspects of control not yet considered, and when they are
considered they make negative feedback control both more accurate and
simpler than compensating control.

This, clearly, is going to be in the new book. So thank you, Bjorn,

Best,

Bill P.

[From
Bjorn Simonsen (2004.08.06,13:45 EuST)]

[From
Bill Powers (2004.08.05.0845 MDT)]

Bjorn Simonsen (2004.08.05,12:18 EuST)–

[From Bjorn Simonsen (2004.08.04,09:30 EuST)]

When I control “knot on the dot” and the “knot is on
the dot” I don’t affect the

environment by any actions. The reference is a1, the perceptual
signal is a1, the

error is zero, the feedback signal is zero (I think) and the
disturbance "knot on

the dot" results in a perceptual signal a1. This is continued
without any affects

till the “knot is not on the dot”

.

[Martin]>Huh? There’s no perceptual signal if the knot is on the dot?

Did I say that? p= ki*ke*ko*e +kd*d.
I meant to say that when „knot is on the dot“, e=0. Then p = kd*d.

I say “huh?” too. What happened to the effect of the output on
qi? The knot is in

the environment, and qo has to affect it continuously to counteract
the continuing

effect of d. If d is a steady pull on one rubber band, qo has to be a

other direction on the other rubber band. The force produced by the output
is

certainly an effect on the environment, isn’t it?

What I
said above is “When I control “knot on the dot” and the “knot is
on the dot” I don’t affect the environment by any actions.” Paraphrased; “When
I wish to hold the “knot on the dot” there is a reference like a1. When I keep “the knot is on the
dot”, the perceptual signal is also like a1. Now, e=zero. I don’t exercise any actions except for the
tension to keep the “knot is not
on the dot”.

That
should give p = kd*d.

I don’t
understand why you ask “What happened to the effect of the output on qi?” The effect of the output is qo = Ko*(r - p).

I
thought that if the “knot is not on the dot” and I wish to keep the “knot
is not on the dot”, then r=p and r-p =0. Now if you go back to >>
(above) p must be p = kd*d.

I have
read what you have written below and I will end this mail with a question and
an example. I have also redone your examples as you asked for.

Let’s analyze the rubber band
experiment. To make the analysis easy, let’s assume

that the rubber bands are identical, so the knot will always be halfway
between

their movable ends. I’ll also ignore the resting length of the rubber
bands, saying

it is zero. Notice that we’re defining all the environmental variables as
positions,

so their values can be added algebraically. This means that we have a much

simplified dependence of the knot position, qi, on the disturbance position
d

and the output position qo.

OK

1. qi = (qo + d)/2

That’s the environment equation.

The controller equations are

p = qi

qo = Ko*(r - p), which reduces to

1. qo = Ko*(r - qi)

We’ll solve for qo by subsituting the right side of equation 1 for qi:

qo = Ko*[r - (qo+d)/2]

Expanding:

qo = Kor - Koqo/2 - Ko*d/2

Collecting terms in qo on the left:

qo + qoKo/2 = Kor - Ko*d/2

Extracting the common term qo on the left:

qo*(1 + Ko/2) = Kor - Kod/2

Dividing both sides by (1 + Ko/2):

``````  Ko*r - Ko*d/2
``````

qo = --------------- .

``````   1 + Ko/2
``````

OK

.

For an exact solution we can subsitute specific values for d and Ko and see

what qo will be. First, however, a value for r is needed. The reference

condition for qi is “knot over dot.” If we place the dot conveniently
at the

zero position of our measurement scale, The desired position of qi is zero,

which is where the dot is, so the reference signal is equal to zero: that

makes the solution even simpler.

Let’s say r = 0, Ko = 100, and d = -20 cm (20 centimeters in the negative
direction)

`````` 0 cm * 100 + 20 cm * 100/2
``````

qo = --------------------------

``````        (1 + 100/2)
``````

qo = 1000/51 cm

qo = 19.61 cm

The position of the knot, qi, will be (by equation 1) the average of qo and d,
or

qi = (19.61 - 20)/2 or

qi = -0.195 cm

So this proportional control system with an output gain of 100 can keep the
knot within 2 mm of the center of the dot when the disturbance is 20 cm, and
even closer when the disturbance is smaller.

OK

Exercise 1: Redo the calculation with an output gain of 2000. What are the
values of qo and qi?

Let’s
say r = 0, Ko = 2000, and d = -20 cm (20 centimeters in the negative direction)

`````` 0 cm * 2000 + 20 cm * 2000/2
``````

qo = --------------------------

``````        (1 + 2000/2)
``````

qo = 2000/1001 cm

qo = 19.98 cm

The position of the knot, qi, will be (by equation 1) the average of qo and d,
or

qi = (19.98 - 20)/2 or

qi = -0.0099 cm

***The higher gain, the greater the output per
unit error. ***

Exercise 2: Redo the calculation with the reference condition set to “knot
5 cm to the right of the dot” (r = 5.00).

Let’s
say r = 5, Ko = 100, and d = -20 cm (20 centimeters in the negative direction)

`````` 5 cm * 100 + 20 cm * 100/2
``````

qo = --------------------------

``````        (1 + 100/2)
``````

qo = 1500/51 cm

qo = 29,41 cm

The position of the knot, qi, will be (by equation 1) the average of qo and d,
or

qi = (29,41 - 20)/2 or

qi = 4.7 cm

If the reference increases (I wish more
strongly to “keep the knot on the dot”), then qi also increases.

Exercise 3: How big is the error when Ko becomes infinite?

Let’s say r = 0, Ko = 1000,000, and d =
-20 cm (20 centimeters in the negative direction)

`````` 5 cm * 1000000 + 20 cm * 1000000/2
``````

qo = --------------------------

``````        (1 + 1000000/2)
``````

qo = 10,000,000/500,001 cm

qo = 19.99 cm

The position of the knot, qi, will be (by equation 1) the average of qo and d,
or

qi = (19.99 - 20)/2 or

qi = 0 cm

If the gain becomes /(almost) infinite,
there are no qi.

Exercise 4: How will this result be affected if the rubber bands are
identical

but their elastic constants are nonlinear (i.e., change with the amount of
stretch)?

Still the perceptual signal will be kept equal to the reference signal,
provided that the loop gain never drops below some minimum value
. qi will no longer numerical be equal the
perceptual signal except at maximum (the normalizing point) (B:CP 276-279).

Exercise 5: Derive the control system equations for the case when the
rubber

bands have different resting lengths and/or different elastic constants Kd
and Ke.

To do this exercise you have to do some physical analysis of the
environment

and the equations will look considerably more complex. The results,
however,

will be the same if you express the output as output position minus resting
length,

and the disturbance as the disturbance position minus resting length.
Assume

that the disturbance position is never less than the resting length of the
associated

rubber band.

I will
do this and come back later. I have noted me your comments.

···

This shows how limited Ashby’s understanding of control systems was. In

the first place, as exercise 3 shows, the error can be made as small as

desired in a negative feedback control system (if it remains stable) simply

by raising the gain. And second of all, the system will remain stable if,
for

any gain, a slowing factor is inserted that reduces the amount of error

correction that takes place per unit time. Such a slowing factor is needed

when there is a transport lag in the system.

May I

As I
wrote to Martin, I define Gain as a value we put in a Simulating model to get
precise control. The higher value in Gain, the greater output per unit error.
(I think I have this from Rick). High Gain means precise control, but systems
with high Gain respond more slowly. The value of Gain and the Slowing factor
must be inversely related.

Different
people wish different wishes in a different degree. Let us have an example where
Mr. A wishes to educate himself to a certain profession. He has a family with
small kids. He looks at his future profession in an economical way, a family
way, from his interests and from the viewpoint of “the best for the state”.
When he controls his perception of future profession, he has activated a most
extensive control system.

Mr. B
wishes to educate to the same profession just because he must have a future
job. When he controls his perception of future profession, he has activated a smaller
control system.

And
now to my question. Is the (p – r) value for Mr. A influenced of a greater Gain
than the same value for Mr. B?

Can we
say that because Mr. A controls a greater control system, it takes longer time before
his p = r?

Here
is my example.

Let’s
say r = 0, Ko = 100, and d = 0 cm (0 centimeters in the negative direction)

`````` 0 cm * 100 + 0 cm * 100/2
``````

qo = --------------------------

``````        (1 + 100/2)
``````

qo = 0/51 cm

qo = 0 cm

The position of the knot, qi, will be (by equation 1) the average of qo and d,
or

qi = (0 - 0)/2 or

qi = 0 cm

If you
go to the top where I wish to keep “the knot on the dot” and if I in a moment
perceive “the knot on the dot”, then p = kd*d.

P = 0
= r

Let’s
say r = 5, Ko = 100, and d = 0 cm (0 centimeters in the negative direction)

`````` 5 cm * 100 + 0 cm * 100/2
``````

qo = --------------------------

``````        (1 + 100/2)
``````

qo = 5/51 cm

qo = 0.098 cm

The position of the knot, qi, will be (by equation 1) the average of qo and d,
or

qi = (00098 - 0)/2 or

qi = 0 cm

If you
go to the top where I wish to keep “the knot on the dot” more strongly and if I
in a moment perceive “the knot on the dot”, then p = kd*d.

P = 0
= r

bjorn

I dont understand why you ask
What happened to the effect of the output on qi? The effect of
the output is qo = Ko*(r - p).
[From Bill Powers (2004.08.06.0944 MDT)]

Bjorn Simonsen
(2004.08.06,13:45 EuST)–

What I said
above is When I control “knot on the dot” and the “knot
is on the dot” I don’t affect the environment by any actions.
Paraphrased; When I wish to hold the “knot on the dot”
there is a reference like a1. When I keep “the knot is on the
dot”, the perceptual signal is also like a1. Now, e=zero. I
dont exercise any actions except for the tension to keep the “knot
is not on the dot”.

OK, I can see the problem now. You are thinking of action as being only a
change of output. You think of a steady output, like holding the end of a
rubber band against a disturbing force, as “no action” if the
hand is not moving. Do I have that right?

In control theory, qo is a measure of the amount of output from the
control system, whether it is constant or changing. When I say
“action” I mean simply the amount of output at any given
instant, however we are measuring it. I do not mean “change of
output.”

That is the output, not the effect of the output. The output is the
position of your hand, the effect of the output is a force applied to the
knot. Notice that if the error is exactly zero in the equations you are
using, the output will be zero. If the output is zero there is nothing to
resist the disturbance because there is no pull opposed to it.

Exercise
1: Redo the calculation with an output gain of 2000. What are the values
of qo and qi?

Let’s say r =
0, Ko = 2000, and d = -20 cm (20 centimeters in the negative direction)

`````` 0 cm * 2000 + 20 cm * 2000/2
``````

qo = --------------------------

``````        (1 +
``````

2000/2)

qo = 2000/1001 cm

This is wrong: 20* 2000/2 = 20,000, not 2000

qo
= 19.98 cm

This is right. A more exact number is 19.98002. You’ll see why I say
that, shortly.

The
position of the knot, qi, will be (by equation 1) the average of qo and
d, or
qi = (19.98 - 20)/2 or
qi = -0.0099 cm

The
higher gain, the greater the output per unit
error.

(A more exact number is qi = -0.0999001)

That is true. Notice that the error is also smaller when the output gain
is higher – but it is not zero. In a proportional control system, the
error is never zero except when the disturbance alone can keep the
perceptual signal equal to the reference signal. Since p = qi, we have p
= 0.0990001.The reference signal is r = 0, so the error (r - p) is (0 -
(-0.0999001)) or

e = 0.0999001. Multiply that error by the gain of 2000 to calculate the
output quantity, and we get

qo = 2000 * 0.0999001 = 19,98002.

Remarkable, no? The amount of error that remains is exactly enough to
produce exactly the amount of output needed to cancel just enough of the
effect of the disturbance to leave exactly that amount of error. So you
see that in a proportional control system, the error can’t go to zero,
with a single exception: when the disturbance has just the right
magnitude. I leave it as an exercise to figure out what that amount of
disturbance is.

Exercise
2: Redo the calculation with the reference condition set to “knot 5
cm to the right of the dot” (r = 5.00).

Let’s say r = 5, Ko = 100, and d = -20 cm (20 centimeters in the negative
direction)
5 cm * 100 + 20 cm * 100/2
qo = --------------------------
(1 +
100/2)
qo = 1500/51 cm
qo = 29,41 cm
The position of the knot, qi, will be (by equation 1) the average of qo
and d, or
qi = (29,41 - 20)/2 or
qi = 4.7 cm

If the reference increases (I wish more strongly to keep the knot
on the dot), then qi also
increases.

This is not the correct interpretation. If the reference is not zero,
then the control system will try to keep the knot at a position away from
the dot. The dot is still at position zero, so when the center of the
knot is exactly on the center of the dot, the knot’s position is also
zero. Here we have the center of the knot being at 4.7 cm, which means
4.7 cm to the right of the dot (with positive measured to the right).
Note that the reference signal was set to 5.0 cm. The knot is 0.29411 cm
to the left of the reference position, and 4.70589 cm away from the
dot.

The reference does NOT determine “how strongly you want to keep the
knot on the dot.” It determines what state of the perception you
want to maintain: on the dot (r = 0) or in this case, 5 cm to the right
of the dot (r = 5). How strongly you react to errors depends on the loop
gain, not on the reference signal.

Exercise 3: How big is the error
when Ko becomes infinite?
Let’s say r = 0, Ko = 1000,000, and d = -20
cm (20 centimeters in the negative direction)
5 cm * 1000000 + 20 cm * 1000000/2
qo = --------------------------
(1 +
1000000/2)
qo = 10,000,000/500,001 cm
qo = 19.99 cm
The position of the knot, qi, will be (by equation 1) the average of qo
and d, or
qi = (19.99 - 20)/2 or
qi = 0 cm

If the gain becomes /(almost) infinite, there are no
qi.

You didn’t answer the question: qi is not the error. The error is the
difference between qi and the position implied by the reference signal,
which in this case is zero, the same position as the dot. Since the dot
is at zero and the reference position is at zero, there is no error and
the error is zero.

Saying “there is no qi” does not mean the same thing as saying
qi = 0.

Exercise
4: How will this result be affected if the rubber bands are identical
but their elastic constants are nonlinear (i.e., change with the
amount of stretch)?

Still
the perceptual signal will be kept equal to the reference signal,
provided that the loop gain never drops below some minimum
value
. qi will no longer numerical be equal
the perceptual signal except at maximum (the normalizing point) (B:CP
276-279).

This is both correct and not correct. You are correct in saying that the
behavior will be the same as with linear rubber bands. If the two rubber
bands are identical, the knot will always be exactly midway between the
free ends even if the rubber bands are nonlinear. So the behavior of the
whole system will be exactly the same.
But it is not correct to say that qi will no longer be equal to the
perceptual signal, because p = qi, always, and that is not changed by
changing the properties of the rubber bands.
It is also not correct to say that the perceptual signal will be kept
equal to the reference signal. In Example 1 above, I showed why
there must be an error, exzctly the right error. If the error were
zero, the output position would be zero and the disturbance from the
other rubber band would cause an error.

DO NOT THINK OF DISTURBANCES, INPUTS, ERRORS, PERCEPTIONS, AND OUTPUTS AS
EVENTS OR CHANGES. THINK OF THEM AS STEADY QUANTITIES OR
MAGNITUDES.

You need differential equations to deal with rates of change as
variables.

Exercise 5: Derive the control
system equations for the case when the rubber

bands have different resting lengths and/or different elastic
constants Kd and Ke.

To do this exercise you have to do some physical analysis of the
environment

and the equations will look considerably more complex. The results,
however,

will be the same if you express the output as output position minus
resting length,

and the disturbance as the disturbance position minus resting length.
Assume

that the disturbance position is never less than the resting length
of the associated

rubber band.

I will do this

As I wrote to
Martin, I define Gain as a value we put in a Simulating model to get
precise control. The higher value in Gain, the greater output per unit
error. (I think I have this from Rick). High Gain means precise control,
but systems with high Gain respond more slowly.

They don’t respond more slowly just because the gain is higher. You have
to MAKE them respond more slowly by changing the slowing factor, if you
want them to remain stable. High gain means precise control, but it also
means that instability is likely. If instability does result from
increasing the gain, you can correct that problem by making the action
slower.

But in algebraic analyses of control systems, there are no slowing
factors because we are not using differential equations. There are no
time delays.

The
value of Gain and the Slowing factor must be inversely
related.

Yes, if you want the highest precision possible. You can always use a
lower gain and leave the slowing factor unchanged, if it doesn’t matter
that the system will correct errors more slowly. Usually you want to use
a slowing factor just sufficient to prevent instability. You don’ usually
want control to be any slower than necessary.

Different
people wish different wishes in a different degree. Let us have an
example where Mr. A wishes to educate himself to a certain profession. He
has a family with small kids. He looks at his future profession in an
economical way, a family way, from his interests and from the viewpoint
of the best for the state. When he controls his perception of future
profession, he has activated a most extensive control
system.

You seem to be picturing this as a single control system. I prefer to say
that one control system controls only one variable. You are speaking of
controlling more than one variable, so more than one control system is
needed. Mr. A is controlling many variables at the same time, so there
are many control systems in operation at the same time, each one still
controlling only one variable.

Mr.
B wishes to educate to the same profession just because he must have a
future job. When he controls his perception of future profession, he has
activated a smaller control system.

Mr.B is controlling fewer variables – only one, in fact. There is only
one control system (that you are considering).

And
now to my question. Is the (p  r) value for Mr. A influenced of a
greater Gain than the same value for Mr. B?

Can we say that because Mr. A controls a greater control system, it takes
longer time before his p = r?

No. All these variables in Mr. A are at about the same level or
organization, I would guess. Each control system has its own gain, which
determines how much the output of that system will change for a unit
change in error signal in that system. Mr. A is controlling several
different perceptions, p1, p2, p3 … relative to several different
reference signals, r1, r2, r3 … Each perception is controlled by a
different control system, independent of the others at this level. The
speed of control in each system is determined by the gain and slowing
factor in each system, which is independent of the others. Since all
these systems operate in parallel, there is nothing about having more
than one system in action that would slow them down, unless they conflict
with each other.

When you talk about different degrees of wishing, I think you’re
confusing wishing for different amounts of something and acting with
different amounts of effort to correct a unit of error. If you want to
earn \$100,000 per year, you are wishing for more income than if you want
to earn \$10,000 per year. You aren’t wishing any harder, you’re just
wishing for a larger amount.But if the amount you actually earn is \$1000
less than that amount you wish to earn, how hard are you willing to work
to bring the amount closer to the amount you want? The man who
earns \$99,000 per year will probably not increase his efforts as much as
the man who earns \$9000 per year, so in fact the man with the lower
wished-for income will try harder than the other man to correct the same
error (can you think of a way to contradict this?). This is a matter of
the loop gain, not the setting of the reference signal. The reference
signal specifies only how much of the perception you will try to
maintain. It doesn’t specify how hard you will try to maintain it when
the perception falls a little short.

I think what people mean by wishing harder is imagining a larger effort
to attain the same wish. You know, you clench your fists and squint your
eyes and squeeze your breath and tense your muscles when you wish harder,
but that has nothing to do with the wishing part – it has to do with
making more effort.

Here
is my example.

Let’s say r = 0, Ko = 100, and d = 0 cm (0 centimeters in the negative
direction)

`````` 0 cm * 100 + 0 cm * 100/2
``````

qo = --------------------------

``````        (1 +
``````

100/2)

qo = 0/51 cm

qo = 0 cm

The position of the knot, qi, will be (by equation 1) the average of qo
and d, or

qi = (0 - 0)/2 or

qi = 0 cm

If you go to the top where I wish to keep the knot on the dot and if I
in a moment perceive the knot on the dot, then p =
kd*d.

P = 0 = r

Yes,in this case the reference position is zero, there is no disturbance,
so no output effort is needed to keep p = r. I don’t know yet what this
is an example of, but I’m trying to follow you.

Let’s
say r = 5, Ko = 100, and d = 0 cm (0 centimeters in the negative
direction)

`````` 5 cm * 100 + 0 cm * 100/2
``````

qo = --------------------------

``````        (1 +
``````

100/2)

qo = 5/51 cm

Mistake here. Should be

qo = 5*100/51 = 500/51 = 9.8 cm.

This means that the next result is wrong, too

qo
= 0.098 cm

The position of the knot, qi, will be (by equation 1) the average of qo
and d, or

qi = (00098 - 0)/2 or

qi = 0 cm

This is wrong even without the above mistake: (0.098 - 0)/ 2 is 0.049,
not 0. But since the 0.098 should be 9.8, the answer should be

qi = (9.8 - 0)/2 = 4.6.

Since p = qi, p = 4.6. The reference value was 5.0, so the error (r-p) is
-0.4 cm. In the second case the error has increased when the reference
position is moved away from the dot. By “reference position” I
mean the position where the knot would be IF the error were zero. But
notice that the knot is now 4.6 cm away from the dot.

Keep trying!

Best,

Bill P.

[From
Bjorn Simonsen (2004.08.08,08:36 EuST)]

[From
Bill Powers (2004.08.06.0944 MDT)]

Bjorn Simonsen (2004.08.06,13:45 EuST)–

What
I said above is “When I control “knot on the dot” and the “knot
is on the dot” I don’t

affect the environment by any actions.” Paraphrased; “When I wish to
hold the "knot on the

dot" there is a reference like a1. When I keep “the
knot is on the dot”, the perceptual signal

is also like a1. Now, e=zero. I
don’t exercise any actions except for the tension to keep the

“knot is not on the dot”.

OK, I can see the problem now. You are thinking of action as being only a
change of output.

You think of a steady output, like holding the end of a rubber band against
a disturbing force,

as “no action” if the hand is not moving. Do I have that right?

Yes and No. We must look at my
Statement in correct relationship. Rohan Luland raised “Rohan Luland
about 2004.08.03, 15:42 EuSt” the question:

it
also raises the question for me, which may be very naive, " do we have to
be able to affect

the environment or the perception in the environment for it to be in a
control loop" are there

some where with either choose to control it or if can not we choose quickly
to control

something else when error is experienced (binary sort of system). it may
not look like a

control loop but still is

I marked the two first lines in
his section, because I am taken up with the idea that we
control all control system we can control
and we perceive all control system
we “just” can perceive/(notice) _ ALL
THE TIME. But we are not conscious/aware
of all these perceptions.”
And I answered “From Bjorn Simonsen (2004.08.04,09:30
EuST)]”

When
I control “knot on the dot” and the “knot is on the dot” I don’t affect the
environment

by any actions. The reference is a1,
the perceptual signal is a1, the error is zero, the feedback

signal is zero (I think) and the disturbance “knot on the dot” results in a
perceptual signal a1.

This is continued without any affects till the “knot is not on the dot”

This was an example when we
" do we do not have to be able to affect the environment or the perception
in the environment for it to be in a control loop". Just that moment. Here
I don’t think about action as being only a change of output. But I think of a steady output, like holding
the end of a rubber band against a disturbing force, as “no action”
if the knot is not moving away from the.

In control theory, qo is a measure of the amount of output from the control
system,

whether it is constant or changing. When I say “action” I mean
simply the amount

of output at any given instant, however we are measuring it. I do not mean
“change of output.”

May I meet this with a comment? An action or behavior is in
response to the presence of an error. The emitted action is transmitted
purposefully, with the intention of changing the state of the world so that my
perception can be made a desired state of goal, which reduces the error signal
to (about) zero. Before my perception can be made a desired state of goal qo
has a value different from zero. At the same time an action is exercised. This
action changes the state of the world and my perception. When my perception
changes, the error and qo changes. When my perception is made a desired state
of goal, qo is (about) zero. At
this time no action is
exercised. When there is no action, the
state of the world doesn’t change, and my perception doesn’t change. Neither does
the error or qo. At this moment when my perceptions have made the desired state
of goal (qo = zero and no actions), I just notice my perceptions, I don’t
control them.

I think there is coherence
between action and “change of output”.

I
don’t understand why you ask “What happened to the effect of the output on
qi?”

The effect of the output is qo = Ko*(r - p).

That is the output, not the effect of the output. The output is the position
of your

hand, the effect of the output is a force applied to the knot. Notice that
if the error

is exactly zero in the equations you are using, the output will be zero. If
the output

is zero there is nothing to resist the disturbance because there is no pull
opposed to it.

Yes you are correct, but I
meant to answer your question a ”What happened to the effect of the output on qi?”
with the answer: “The effect of the output is zero because qo = Ko*(r - p) AND
(r-p) = zero.

NOW READ THE NEXT SECTION TWICE.

In an ordinary rubber band game
it happens that the knot is on the dot, but using many decimals, qo is never
ZERO.

If you and I play a SPECIAL
rubber band game where we agree upon the dot, put the knot over the dot and
keep it so. Then we say “if you stretch the rubber band I shall try to keep
“knot over the dot”. The time before
you stretch the rubber band the knot is over the dot, I stretch with just the
same power as you in opposite direction (linear output function). NOW p=r, e=0,
qo =0, p=0. Controlling means***: A is said to control B IF for every disturbing influence acting
on B, A generates an action that tends strongly to counteract the effect of the
disturbing influence on B.***

NOW we have an example where a
perception is just noticed and not controlled. Have I learned? If I have learned,
my comment to Rohan Luland was wrong. Because in my example, no control took
place.

Exercise
2: Redo the calculation with the reference condition set to “knot 5 cm to
the right of the dot” (r = 5.00).

Let’s say r = 5, Ko = 100, and d = -20 cm (20 centimeters in the negative
direction)
5 cm * 100 + 20 cm * 100/2
qo = --------------------------
(1 + 100/2)
qo = 1500/51 cm
qo = 29,41 cm
The position of the knot, qi, will be (by equation 1) the average of qo and d,
or
qi = (29,41 - 20)/2 or
qi = 4.7 cm

If the reference
increases (I wish more strongly to “keep the knot on the dot”), then qi
also increases.

This is not the correct interpretation. If the reference is not zero, then
the control system

will try to keep the knot at a position away from the dot. The dot is still
at position zero,

so when the center of the knot is exactly on the center of the dot, the
knot’s position is

also zero. Here we have the center of the knot being at 4.7 cm, which means
4.7 cm to the

right of the dot (with positive measured to the right). Note that the
reference signal was

set to 5.0 cm. The knot is 0.29411 cm to the left of the reference
position, and 4.70589 cm

away from the dot.

The reference does NOT determine “how strongly you want to keep the
knot on the dot.”

It determines what state of the
perception you want to maintain: on the dot (r = 0) or in this

case, 5 cm to the right of the dot (r = 5). How strongly you react to
errors depends on the

loop gain, not on the reference signal.

Here I am learning aspects of
PCT. The way I understand your comments is that you think upon a linear input
function where the knot is moving along a “horizontal” line through the dot.
The game doesn’t tolerate movements orthogonal this line ((r=5) is 5 cm to the
right of the dot). When orthogonal movements are actual, we talk about a
nonlinear input function.

Exercise
3: How big is the error when Ko becomes infinite?

Let’s
say r = 0, Ko = 1000,000, and d = -20 cm (20 centimeters in the negative
direction)
0 cm * 1000000 + 20 cm * 1000000/2
qo = --------------------------
(1 +
1000000/2)
qo = 10,000,000/500,001 cm
qo = 19.99 cm
The position of the knot, qi, will be (by equation 1) the average of qo and d,
or
qi = (19.99 - 20)/2 or
qi = 0 cm

If the gain
becomes /(almost) infinite, there are no qi.

You didn’t answer the question: qi is not the error. The error is the
difference between

qi and the position implied by the reference signal, which in this case is
zero, the same

position as the dot. Since the dot is at zero and the reference position is
at zero, there

is no error and the error is zero.

Saying “there is no qi” does not mean the same thing as saying qi
= 0.

``````            e= qi – r

qo =
``````

10,000,000/500,001 = 19.99996

``````            qi =  (19.99996 –20) = - 0.00004

e = -0.00004 – 0 = -0.00004

***If the gain becomes
``````

/(almost) infinite, the error becomes still smaller than when the gain is
great.***

………………. The reference signal specifies only how much of the perception you

will try to maintain. It doesn’t specify how hard you will try to maintain
it when the

perception falls a little short.

Thank you.

You have got an insight in my
way of thinking when you wrote:

DO
NOT THINK OF DISTURBANCES, INPUTS, ERRORS, PERCEPTIONS, AND

OUTPUTS AS EVENTS OR CHANGES. THINK OF THEM AS STEADY

QUANTITIES OR MAGNITUDES

I will.

bjorn

Re: Knot on dot [Was Perceived Present [was Shared
referen
[Martin Taylor 2004.08.08.0927]

I’m away for about 3 weeks starting in a few hours, so this has
to be quick.

[From Bjorn Simonsen (2004.08.08,08:36
EuST)]

[From Bill Powers (2004.08.06.0944
MDT)]

Bjorn Simonsen (2004.08.06,13:45
EuST)–

OK, I can see
the problem now. You are thinking of action as being only a change of
output.

You think of a steady output, like holding the end of a rubber
band against a disturbing force,

as “no action” if the hand is not moving. Do I have that
right?

Yes and No. We must look at my Statement in correct
relationship. Rohan Luland raised “Rohan Luland about 2004.08.03,
15:42 EuSt” the question:

it also raises the question for me, which may be
very naive, " do we have to be able to affect

the environment or the perception in the environment for it to be
in a control loop" are there

some where with either choose to control it or if can not we
choose quickly to control

something else when error is experienced (binary sort of system).
it may not look like a

control loop but still is

I marked the two first lines in his
section,

As you should, becasue what it asks is “Is there any way of
affecting our perception other than by having an influence on the
environment?” It involves the notion of controlling in
imagination, and whether our imagination affects what we perceive from
the environment.

But nothing of that shows up in the rest of your message.

I don’t think about action as being only a
change of output. But I think of a steady output, like holding the end
of a rubber band against a disturbing force, as “no action”
if the knot is not moving away from the.

You should be VERY careful not to treat a variable that has a
value of zero as the same as the absence of the variable. “x=0”
is a very different statement from “there is no such thng as
‘x’”.

In control theory, qo is a measure of the amount of output from
the control system,

whether it is constant or changing. When I say “action”
I mean simply the amount

of output at any given instant, however we are measuring it. I do
not mean “change of output.”
May I meet this with a comment? An action or behavior
is in response to the presence of an error.

No. Same mistake. It is a function of the magnitude of the error
and of the history of the error. In a control loop “the presence
of an error” is guaranteed by the “wiring” of the
control loop. A comparator exists, and has values at its inputs.
Therefore it has a value at its output, which is labelled “the
error”.

-----------------a different
problem------------------------------

NOW READ THE NEXT SECTION TWICE.
In an ordinary rubber band game it happens that the
knot is on the dot, but using many decimals, qo is never
ZERO.
If you and I play a SPECIAL rubber band game where we
agree upon the dot, put the knot over the dot and keep it so. Then we
say "if you stretch the rubber band I shall try to keep “knot over
the dot”. The time before you stretch the rubber band the
knot is over the dot, I stretch with just the same power as you in
opposite direction (linear output function). NOW p=r, e=0, qo =0, p=0.
Controlling means***: A is said to control B IF for every
disturbing influence acting on B, A generates an action that tends
strongly to counteract the effect of the disturbing influence on
B.***
NOW we have an example where a perception is just
noticed and not controlled. Have I learned? If I have learned, my
comment to Rohan Luland was wrong. Because in my example, no control
took place.

You are still controlling a perception, but it is a perception
that your action matches what the other is doing, not of the position
of the knot. The position of the knot now is a side-effect of your
control for matching the other’s actions.

----------------problem 3--------------------

Here I am learning aspects of PCT. The way I
function where the knot is moving along a “horizontal” line
through the dot. The game doesn’t tolerate movements orthogonal this
line ((r=5) is 5 cm to the right of the dot). When orthogonal
movements are actual, we talk about a nonlinear input
function.

No we don’t. It’s
still linear, but two perceptions are being controlled independently
by two different control systems, one for position aling the line of
the rubber band, one for position orthogonal to that line.
“Non-linear” means something quite different, and
potentially could apply separately to each of those control systems.
“Non-linear” shows up in the equations for the control
system itself. For example, instead of q0 = ke, you might have q0 = k

• log(e) or q0 = k * e^2. In human perception, you often have
something along the lines of p = log(s) or p = s^p.

Martin

Yes and No. We must look at my
Statement in correct relationship. Rohan Luland raised Rohan
Luland about 2004.08.03, 15:42 EuSt the question:
When I control knot on the dot
and the knot is on the dot I dont affect the environment

by any actions. The reference is a1, the perceptual signal is a1,
the error is zero, the feedback

signal is zero (I think) and the disturbance knot on the dot results in
a perceptual signal a1.

This is continued without any affects till the knot is not on the
dot
[From Bill Powers (2004.08.07.0642 MDT)]

Bjorn Simonsen
(2004.08.08,08:36 EuST)–

This
was an example when we " do we do not have to be able to affect the
environment or the perception in the environment for it to be in a
control loop". Just that moment. Here I dont think about action as
being only a change of output. But I think of a steady output, like
holding the end of a rubber band against a disturbing force, as “no
action” if the knot is not moving away from
the.

I don’t know of any case where we do not have to be able to affect the
environment or the perception for it to be a control loop. When you say
“be able to affect the environment”, I understand that this
means having an output device (an arm and a hand) connected to the
environment (the finger of the hand hooked through the rubber band) so
that the output device can affect the controlled quantity, the position
of the knot. If I removed the rubber band from your finger, you would no
longer be able to control the position of the knot, although you could
still see it. I don’t see how you could continue to control the position
of the knot if you could no longer affect its position. Please
explain.

When
my perception changes, the error and qo changes. When my perception is
made a desired state of goal, qo is (about) zero. At this time
no action is exercised. When there is no action, the state
of the world doesnt change, and my perception doesnt change. Neither
does the error or qo. At this moment when my perceptions have made the
desired state of goal (qo = zero and no actions), I just notice my
perceptions, I dont control them. think there is coherence between
action and change of output.

You are defining control in too limited a way, so that one particular
state of an active control system is defined as “not
controlling.” But what I am trying to get across is that that in
most cases, that state of small error cannot be maintained against a
steady disturbance unless the control system is producing a continuous
and NON-ZERO output. You still seem to be assuming that if there is no
change in the output, there is no output. That is not
true.

I
dont understand why you ask What happened to the effect of the output
on qi?

The effect of the output is qo = Ko*(r - p).

That is the output, not the effect of the output. The output is the
position of your

hand, the effect of the output is a force applied to the knot. Notice
that if the error

is exactly zero in the equations you are using, the output will be
zero. If the output

is zero there is nothing to resist the disturbance because there is
no pull opposed to it.

Yes
a What happened to the effect of the output
on qi? with the answer: The effect of the output is zero because
qo = Ko*(r - p) AND (r-p) = zero.

No, (r-p) is NOT zero. It is a small amount, which is then amplified by
the gain in the output function (Ko) to put the hand in the right
position to counteract the effect of the disturbance on the knot. If the
error became larger, the hand would move farther from the zero position.
That would move the knot off the dot even further, causing the error to
increase, so the hand would move back toward the dot again. Except that
this sequential description is wrong: the hand simply moves to the
position where the knot is just far enough off the dot (not very far) for
the resulting error to keep the hand where it is. Control continues
whether the hand and knot are moving or stationary.

NOW

In an ordinary rubber band game it happens that the knot is on the dot,
but using many decimals, qo is never ZERO.
NOW we have an example
where a perception is just noticed and not controlled. Have I learned? If
I have learned, my comment to Rohan Luland was wrong. Because in my
example, no control took place.

I read that twice and it is still misleading. You don’t need “many
decimals” to show that qo is not zero when the knot is on the dot.
You made some arithmetic mistakes that made this seem to be true, but in
your example, the output was actually something like 9 units, not zero,
even omitting the decimal places.
Let’s diagram the rubber band experiment (this will give some hints about
how to analyze the more general case in one of the Exercises). Let’s
start with the rubber bands at their resting lengths, the knot over the
dot, and both the experimenter’s hand (He) and the controller’s hand (Hc)
holding the ends of the rubber bands. Under these conditions we can that
He = 0, Hc = 0, and qi = 0, where qi is the position of the knot. The dot
always remains at zero.

5…4…3…2…1…0…1…2…3…4…5

He=============8==============Hc
-5…-4…-3…-2…-1…-0
knot
0…1…2…3…4…5

x

dot
The scale on the top applies to the knot (the figure 8) and dot (the x)
only
Now consider what happens when the disturbance, Hd, changes from 0 to -3
units. First, let’s see what would happen to the knot if the gain of the
control system was zero.

-5…-4…-3…-2…-1…-0…1…2…3…4…5

`````` (d) Hd===================8===================Hc
``````

(qo)
-5…-4…-3…-2…-1…-0
knot(qi)
0…1…2…3…4…5

x

dot
We have a disturbance d of -3 units and an output quantity Hc of 0 units,
so the position of the knot is (He + Hc)/2 = (-3 + 0)/2 = -1.5 units. The
8, which is the knot, is at the position -1.5 on its scale. The reference
signal is set to zero and p = qi (qi is the knot position), so the error
signal will have a value of (r - p) = [0 - (-1.5)] = +1.5
units.
If the gain Ko is greater than zero, the error signal will produce
output, and Hc will be farther to the right. Suppose the gain is 1. In
that case the output will be
qo = Kor - (Kod/2)/(1 + Ko/2) or
qo = 0 - (1*(-3)/2)/(1+1/2) or
qo = 1
The diagram will look like this:

-5…-4…-3…-2…-1…-0…1…2…3…4…5

Hd=====================8======================Hc
-5…-4…-3…-2…-1…-0
knot
0…1…2…3…4…5

x

dot
The theoretical knot position is -1, which is what we see.
Now if we let the gain be 100, the error will become so small that it
looks like zero in this diagram. The output quantity will be further
still to the right, and we will see this:

-5…-4…-3…-2…-1…-0…1…2…3…4…5

Hd=========================8==========================Hc
-5…-4…-3…-2…-1…-0
knot
0…1…2…3…4…5

x

dot
Now the knot is over the dot as nearly as we can see on this scale. The
actual value of qo will be 2.941176471 units (to the right) and the knot
position will be -0.02941164 units (to the left). The error signal will
be positive by the same amount since p = qi and e = r - p. The reference
signal r is zero, Ko is 100, so the output quantity Hc (otherwise known
as qo) will be 2.941164 units as nearly as my calculator can show. That’s
the same number we started with so we know the calculation is right. The
rounding error in this result is 0.00000071, which is just because my
calculator displays only 12 characters and rounds the result.

If you and I
play a SPECIAL rubber band game where we agree upon the dot, put the knot
over the dot and keep it so. Then we say if you stretch the rubber band
I shall try to keep knot over the dot. The time before you
stretch the rubber band the knot is over the dot, I stretch with just the
same power as you in opposite direction (linear output function). NOW
p=r, e=0, qo =0, p=0. Controlling means***: A is said to control B IF
for every disturbing influence acting on B, A generates an action that
tends strongly to counteract the effect of the disturbing influence on
B.***

Sorry, no. You are controlling even if the output is constant, and even
if it is zero. You would not be controlling if the loop were broken,
which is a different matter. Suppose that with the rubber bands just
relaxed at zero, I quietly cut your end of the rubber bands while you’re
not looking, and arrange the cut ends so they still look joined. Now,
even if you think you have control, you do not, because the environment
equation has changed. It is no longer qi = (d + qo)/2, but qi = d (using
the diagram above). Now you have no control. Your output no longer has
any effect on qi. The effect of qo on qi in our equation is zero, so it
doesn’t matter how large or small qo is. It still has no effect, and
there is no control.

Exercise
2: Redo the calculation with the reference condition set to “knot 5
cm to the right of the dot” (r = 5.00).

This is not the
correct interpretation. If the reference is not zero, then the control
system

will try to keep the knot at a position away from the dot. The dot is
still at position zero,

so when the center of the knot is exactly on the center of the dot,
the knot’s position is

also zero. Here we have the center of the knot being at 4.7 cm, which
means 4.7 cm to the

right of the dot (with positive measured to the right). Note that the
reference signal was

set to 5.0 cm. The knot is 0.29411 cm to the left of the reference
position, and 4.70589 cm

away from the dot.

The reference does NOT determine “how strongly you want to keep
the knot on the dot.”

It determines what state of the perception you want to
maintain: on the dot (r = 0) or in this

case, 5 cm to the right of the dot (r = 5). How strongly you react to
errors depends on the

loop gain, not on the reference signal.

Here I am learning aspects of PCT. The way I understand your comments is
that you think upon a linear input function where the knot is moving
along a horizontal line through the dot. The game doesnt tolerate
movements orthogonal this line ((r=5) is 5 cm to the right of the dot).
When orthogonal movements are actual, we talk about a nonlinear input
function.

No, we just add a second control system that controls the knot position
in the y direction, independently of the first one. The new qi is
controlled by a new qo: a position of the end of the rubber band at right
angles to the line of the rubber bands.

But you didn’t comment on my demonstration that your calculation was
wrong. Increasing the reference signal does not make the system try
harder; it changes the location where the knot is to be maintained,
moving it to the right.

Exercise 3: How big is
the error when Ko becomes infinite?

If
the gain becomes /(almost) infinite, there are no qi.

You didn’t answer the question: qi is not the error. The error is the
difference between

qi and the position implied by the reference signal, which in this
case is zero, the same

position as the dot. Since the dot is at zero and the reference
position is at zero, there

is no error and the error is zero.

Saying “there is no qi” does not mean the same thing as
saying qi = 0.

above.

I think you missed my meaning: the problem may be one of translation into
English. When you say “There is no qi” this is taken to mean
“There is no variable that is being controlled.” or “there
is no knot.” That is not the same as saying “There is a
variable that is being controlled, and value at which it is being
controlled is zero.”

I hope that this is helping.

Best,

Bill P.