[From Alex Gomez-Marin (2016.09.14.1826)]
AGM: Rick, we are where we were 2 months ago when I posted my initial question. I DON’T have a generative model for the speed-curvature power law. That is why I posed the question. But you don’t have it either. All you have is basic ad-hoc reference trajectories as a function of time that you inject in your so-called demo, which is barely more than a leaky integrator of such reference trajectories, and then a plot for V and A that approaches a power law. With that simple case you cannot prove that the power law is a trivial result, nor an obligatory result, nor valid or invalid for all other infinitely possible ad hoc reference trajectories one may inject. Plus, in all the papers I have sent you about the power law, researchers have obviously studied ellipses and so forth, and gotten much further than you. So, you are not reinventing the wheel there with your demo. Finally, your demo is not a PCT demo in the sense that we don’t know what controlled perceptual variables GIVE RISE to the output. So, I guess, we are still in the “behavioral illusion” phase. That is not to mention any comments on your infamous re-writing of the mathematical equation for curvature. To sum up, you explained nothing and, instead, created a lot of stress, confusion and waste of time in the net. And you still do.
···
On Wed, Sep 14, 2016 at 5:48 PM, Richard Marken rsmarken@gmail.com wrote:
[From Rick Marken (2016.09.14.0845)]
On Wed, Sep 14, 2016 at 1:52 AM, Alex Gomez-Marin agomezmarin@gmail.com wrote:
AGM: In other words, join me in the efforts of these newly created and much needed ONG:
(RCP = Rhetoric Control Theory)
RM: I presume you are implying that my PCT explanation of the power law is all “rhetoric”. That strikes me as odd since I have produced a PCT model, tested it against data and showed how it accounts for the power law; while all the “non-rhetorical” PCT side has produced is, well, rhetoric (mainly having to do with how my math is wrong and that the power law “suggests” the existence of some unspecified controlled variable). Until I see your PCT model that accounts for the existence of the power I think all that needs to be posted at the StopRCP site is the arguments of the ostensibly non-rhetorical PCT side of the discussion.
Best
Rick
–
Richard S. Marken
“The childhood of the human race is far from over. We
have a long way to go before most people will understand that what they do for
others is just as important to their well-being as what they do for
themselves.” – William T. Powers
On Wed, Sep 14, 2016 at 8:38 AM, Alex Gomez-Marin agomezmarin@gmail.com wrote:
martin, the csgNet is now a hybrid of a basic math accademy with an entertainment program for writing and joking with those who write back. stay tuned!
On Wednesday, 14 September 2016, Martin Taylor mmt-csg@mmtaylor.net wrote:
[Martin Taylor 2016.09.13.22.49]
[From Rick Marken (2016.09.13.1610)]
If you actually read and thought about what you quote, I cannot read
your question as anything other than a not too subtle joke, so I
have to assume that you either did not read it or the maths, which I
tried to explain at the most basic level O could manage without
seeming to insult you (which I was afraid I was doing anyway), was a
bit too deep for you.Isuggest you try again, concentrating on figuring out why this DOES
explain why you have been making the same kind of mistake as the
“divide by zero” error. Perhaps I should repeat the last lines: " When we put all this together, we come to
the way this is a variant of the “divide by zero” error. That
error depends on the fact that you can put any variable at all in
for “x” in “x/0 = infinity”. The – shall we call it – the
“curvature error” depends on the fact that you can use anything at
all for V (including the measured values), provided only that V is
defined as ds/dz where z is some variable for which ds/dz exists
everywhere. You therefore cannot use the curvature equation in any
way to determine V."I emphasize "IN ANY WAY". Since I may be a little too subtle when
I say this, it simply says that your equation V = D1/3*C1/3 means
nothing at all, because it is true when V is any variable at all
that satisfies a very loose condition. If it is a velocity it can
be any velocity at all, or it can be any value of any variable
(and here I will repeat myself) that depends on any other variable
“z” whatever for which ds/dz is everywhere calculable.
G+O showed the correct formulas, and I said so. You did the correct
FORMAL algebra. Your math error was and apparently continues to be
the equivalent of the “divide by zero” error, which also depends on
doing the algebra correctly. The “divide by zero” or its “curvature
error” equivalent is a math error if ever there was one. It could be
and should be easily correctable, but apparently it isn’t. I don’t
know why.Martin
RM: Oh, I see. It’s contained in this:
MT: Now we have to see how
they came to equation (9). That’s a bit more
complicated, so please bear with me.MT: They presumably
either used someone else’s derivation or made their
own, starting from one of several equivalent
measures of curvature, one of which is C = 1/R where
R is the radius of the osculating circle at the
point of concern. Another one is developed using
vector calculus, which I have no intention of
introducing into this discussion. It is C = dx/dsd2y/ds2 - dy/ds * d2x/ds2 , where s is distance along
the curve from some arbitrary starting point.
For G+O this formula
was not very convenient, because they would have had
to measure these first and second derivatives of x
and y with respect to distance along the curve
fairly accurately. But they had a trick available,
in the “chain rule” of differentiation: dx/dydy/dz
= dx/dz. The "dy"s cancel out just like ordinary
variables. Using the chain rule on the first
derivative gives you the rule for the second
derivative, and so on. For the second derivative the
rule is (d2x/dy2)(dy/dz)2 = d2x/dz2.
Using the chain rule,
G+O could multiply the formula for C by (ds/dz}3/(ds/dz)3 = 1, for any variable z
that allowed the differentiation, to get C =
((dx/ds)(ds/dz)(d2y/ds2)(ds/dz)2)(ds/dz)3 - (dy/ds)(ds/dz)(d2x/ds2)__(ds/dz)__3)/(ds/dz)3 . This formula is true
(allowing for typos) for variable “z” whatever (as
with the divide by zero example), but it wouldn’t
have helped G+O very much, had it not been that for
one particular variable they already had measures
they could use. Those measures were the ds/dt
velocity and the derived d2s/dt2 values they had obtained
from their observations of movement. Using those
measures, they could set “z” = t (time), making
dx/dt = dx/ds*ds/dt. They could then take advantage
of their measured velocities to substitute for
ds/dt, and writeC = (dx/dt*d<sup>2</sup>y/dt<sup>2</sup>)/V<sup>3</sup> - (dy/dt*d<sup>2</sup>y/dt<sup>2</sup>)/V<sup>3</sup> Oh goody! We don't have
to measure anything new to get our curvatures. We
can use the values of dx/dt and dy/dt that we got
before! Very handy. … But also very confusing,
because it made the published equations look as
though the V3/V3 multiplier was special to
the velocities they measured, whereas it was simply
a convenient choice from a literally infinite
variety of choices they could have made. G+O made it
even more confusing in the publication by using the
Newton dotty notation, which made it look as though
there was something necessary about the time
differentiation in the curvature equation.When we put all this
together, we come to the way this is a variant of
the “divide by zero” error. That error depends on
the fact that you can put any variable at all in for
“x” in “x/0 = infinity”. The – shall we call it –
the “curvature error” depends on the fact that you
can use anything at all for V (including the
measured values), provided only that V is defined as
ds/dz where z is some variable for which ds/dz
exists everywhere. You therefore cannot use the
curvature equation in any way to determine V.Does this "Sunday" explanation help?
RM: Not really. Are you saying that G+O used the
wrong formulas for V and R? Or that the formulas they
published are not actually the ones they used to
compute V and R? Or that there is no way to compute R
since we can’t measure ds? Either way, you can’t say I
made a math error since I did the math correctly on
the formulas I was given. (And, as I mentioned, the
results came out exactly right).
MT: but the critical
point of my suggestion that you read my
earlier message is contained in this:
RM: Contained in what?
RM: Rather than saying that I was making a math
error, it would have helped if you had just said:
“these are the correct formulas for computing V and R”
and showed me the formulas.
That would have saved a lot of trouble. So how
about it; what are the correct formulas for computing
V and R?
Best
Rick
–
Richard S. Marken
"The childhood of the human
race is far from over. We
have a long way to go before
most people will understand
that what they do for
others is just as important to
their well-being as what they
do for
themselves." – William T.
Powers
–
Richard S.
Marken
"The childhood of
the human race is
far from over. We
have a long way to
go before most
people will
understand that what
they do for
others is just as
important to their
well-being as what
they do for
themselves." –
William T. Powers