A New (and Astonishing) Power-Law Relationship

[From Bruce Abbott (2018.01.01.1145 EST)]

I have discovered a heretofore unsuspected relationship between the area of a circle and the speed with which the circumference of the circle is traced.

A point traces the circumference of a circle at velocity V, completing the transit in time T. Then the area A of the circle is given by

A = V2T2/4π

Solving for V:

V = 2A1/2π1/2/T

Notice that the velocity with which the circle is traced is a power-law function of the circle’s area, with an exponent of 1/2. Converting to logarithms:

Log V = log 2 + ½ log A + ½ log π – log T

Where it appears that velocity of the circumscribing point and the area of the circle are independent, this is only due to the influence of log T, which allows the relationship to depart from what otherwise would be a mathematically enforced power relationship.

It amazes me that professional mathematicians have overlooked this important fact!

Bruce

From Fred Nickols 2018.04.01.1201 ET

Hmm. Today must be April 1st.

Fred Nickols

···

On Apr 1, 2018, at 11:47 AM, Bruce Abbott bbabbott@frontier.com wrote:

[From Bruce Abbott (2018.01.01.1145 EST)]

I have discovered a heretofore unsuspected relationship between the area of a circle and the speed with which the circumference of the circle is traced.

A point traces the circumference of a circle at velocity V, completing the transit in time T. Then the area A of the circle is given by

A = V2T2/4Ï€

Solving for V:

V = 2A1/2Ï€1/2/T

Notice that the velocity with which the circle is traced is a power-law function of the circleâ€™s area, with an exponent of 1/2. Converting to logarithms:

Log V = log 2 + Â½ log A + Â½ log Ï€ –“ log T

Where it appears that velocity of the circumscribing point and the area of the circle are independent, this is only due to the influence of log T, which allows the relationship to depart from what otherwise would be a mathematically enforced power relationship.

It amazes me that professional mathematicians have overlooked this important fact!

Bruce

[From Bruce Abbott (2018.01.01.1145 EST)]

Â

``````      I have discovered a heretofore unsuspected
``````

relationship between the area of a circle and the speed with
which the circumference of the circle is traced.

Â

``````      A point traces the circumference of a
``````

circle at velocity V, completing the transit in time T.Â Then
the area A of the circle is given by

Â

A = V2T2/4Ï€

Â

Solving for V:

V = 2A1/2Ï€1/2/T

Â

``````      Notice that the velocity with which the
``````

circle is traced is a power-law function of the circleâ€™s area,
with an exponent of 1/2.Â Converting to logarithms:

Â

Log V = log 2 + Â½ log A + Â½ log Ï€ – log T

Â

Where it appears that velocity of
the circumscribing point and the area of the circle are
independent, this is only due to the influence of log T, which
allows the relationship to depart from what otherwise would be
a mathematically enforced power relationship.

Â

``````      It amazes me that professional
``````

mathematicians have overlooked this important fact!

Â

Bruce

Â

[From Bruce Abbott (2018.04.01. 1300 EST)]

[Martin Taylor 2018.04.01.12.45]

[From Bruce Abbott (2018.01.01.1145 EST)]

I have discovered a heretofore unsuspected relationship between the area of a circle and the speed with which the circumference of the circle is traced.

A point traces the circumference of a circle at velocity V, completing the transit in time T. Then the area A of the circle is given by

A = V2T2/4Ï€

Solving for V:

V = 2A1/2Ï€1/2/T

Notice that the velocity with which the circle is traced is a power-law function of the circleâ€™s area, with an exponent of 1/2. Converting to logarithms:

Log V = log 2 + Â½ log A + Â½ log Ï€ – log T

Where it appears that velocity of the circumscribing point and the area of the circle are independent, this is only due to the influence of log T, which allows the relationship to depart from what otherwise would be a mathematically enforced power relationship.

It amazes me that professional mathematicians have overlooked this important fact!

Bruce

Your analysis ignores the effects of gravity on the velocity. The potential energy gained on the way up to the top must be released on the way down the other side. Your analysis has the velocity equal all the way round, but actually it must be faster on the way down than on the way up. You have to incorporate Einstein’s equations for the curvature of space to the equations that in their simplicity treat only the curvature of a Euclidean circle.

Martin

I hadnâ€™t thought of that. Â I guess I implicitly assumed that the circle was being traced on a horizontal surface.Â But if we are going to bring in the curvature of space, I suppose we should bring relativistic time-dilation into the picture as well, assuming that the point is tracing the circle at some substantial fraction of the speed of light.

Bruce

kepler would be proud of you

···

Alex Gomez-Marin, PhD

Instituto de Neurociencias

behavior-of-organisms.org

oh, no… i took the email seriously!, which demonstrates what counts as surprise in csgNet for me now… buff…

···

Alex Gomez-Marin, PhD

Instituto de Neurociencias

behavior-of-organisms.org

Alex Gomez-Marin, PhD

Instituto de Neurociencias

behavior-of-organisms.org

[From Bruce Abbott (2018.04.02.1725 EST)]

For those of you who may have been taken in by my post (reproduced at bottom below), April Fool!

But there was a serious purpose in posting it. I had presented the following equation:

V = 2A1/2π1/2/T

This form parallels Rick Marken’s analysis (Marken & Shaffer, 2017) of the formula for computing the radius of curvature at a given point as the point traces some curved path. That formula is as follows:

R = V3/D, where V is the velocity of the point along the curve and D is computed from the velocities and accelerations of the point along the X and Y directions. Solving for V yields:

V = R1/3/D1/3, which is similar in form to my equation above relating velocity to circle area.

Converted to log form, my equation becomes:

log V = log 2 + ½ log A + ½ log π – log T

And Rick’s becomes:

log V = 1/3 log R - 1/3 log D

Empirically, a power-law relationship is found between V and R. This is obtained by doing a linear regression analysis in which log R is the predictor variable and log V is the criterion variable:

log V = a + b*log R,

where a is the intercept of the best fitting line relating log V to log R and b is the slope. Empirically, b has often been found to be close to 1/3, which would be the power exponent of R when the equation is expressed in its original (non-log) form. However, empirically the exponent can and often does depart from 1/3.

Rick notes that this regression analysis omits log D as a predictor variable. He asserts that it is this omission that allows the exponent to depart from 1/3, and that if log D is included, then the mathematics of the calculation of the radius of curvature necessarily impose a 1/3 exponent in the formula relating V to R.

In reaching this conclusion, Rick has fallen into a mathematical fallacy, because the radius of curvature itself depends on D. It is the division of V3 by D that makes R a pure measure of distance, uninfluenced by the velocity of the point that moves along the curve.

To better understand this assertion, consider my parallel construction “proving” that the velocity of a point traversing a circle depends on the circle’s area. Obviously this is nonsense – you can go around the circle at any velocity whatsoever no matter what the area of the circle may be. Yet my equation seems to indicate that the velocity of the point is mathematically determined by the circle’s area. How can this be? Well, let’s have a look at my original equation again:

V = 2A1/2π1/2/T

The key is to understand the role of T. Let’s move T to the other side of the equation:

VT = 2A1/2π1/2

V is the velocity of the point in, say, cm/second. T is the time in seconds required to traverse the circle at this particular velocity

V cm/s times T s = cm, a distance. In fact it is the distance around the circle ordinarily referred to as the circle’s circumference. I’ll call it C.

So the formula can now be rewritten as

C = 2A1/2π1/2.

I think you will agree that the area of a circle depends on its circumference. The two measured variables, V and T, were used as a means of determining the circumference. The faster you go around the circle, the sooner you will complete the circuit, so the two variables vary inversely in such a way that the circumference that you determine from V and T will be the same for a given circle no matter what speed you go around it at.

By exactly the same logic, in the formula for the radius of curvature, the denominator D of the formula compensates for the different velocities at which the point may be moving along the curve, so that when V3 is divided by D, the result, the radius, is purely a measure of distance that will be the same value no mater what the velocity of the point was observed to be when the curve was drawn. Because R as computed is independent of the velocity used to compute it, the empirically observed relationship between V and R could be any relationship whatever, and the empirically observed power-law relationship is not an artifact of the way in which R is computed.

It’s as simple as that.

Bruce

[From Bruce Abbott (2018.01.01.1145 EST)]

I have discovered a heretofore unsuspected relationship between the area of a circle and the speed with which the circumference of the circle is traced.

A point traces the circumference of a circle at velocity V, completing the transit in time T. Then the area A of the circle is given by

A = V2T2/4π

Solving for V:

V = 2A1/2π1/2/T

Notice that the velocity with which the circle is traced is a power-law function of the circle’s area, with an exponent of 1/2. Converting to logarithms:

Log V = log 2 + ½ log A + ½ log π – log T

Where it appears that velocity of the circumscribing point and the area of the circle are independent, this is only due to the influence of log T, which allows the relationship to depart from what otherwise would be a mathematically enforced power relationship.

It amazes me that professional mathematicians have overlooked this important fact!

Bruce

[From Bruce Abbott (2018.04.02.1735 EST)]

Correction: The superscript code somehow got lost in the formula for R: it reads R = V3/D but should read V3/D. I have corrected it in the text below.

Bruce

[From Bruce Abbott (2018.04.02.1725 EST)]

For those of you who may have been taken in by my post (reproduced at bottom below), April Fool!

But there was a serious purpose in posting it. I had presented the following equation:

V = 2A1/2π1/2/T

This form parallels Rick Marken’s analysis (Marken & Shaffer, 2017) of the formula for computing the radius of curvature at a given point as the point traces some curved path. That formula is as follows:

R = V3/D, where V is the velocity of the point along the curve and D is computed from the velocities and accelerations of the point along the X and Y directions. Solving for V yields:

V = R1/3/D1/3, which is similar in form to my equation above relating velocity to circle area.

Converted to log form, my equation becomes:

log V = log 2 + ½ log A + ½ log π – log T

And Rick’s becomes:

log V = 1/3 log R - 1/3 log D

Empirically, a power-law relationship is found between V and R. This is obtained by doing a linear regression analysis in which log R is the predictor variable and log V is the criterion variable:

log V = a + b*log R,

where a is the intercept of the best fitting line relating log V to log R and b is the slope. Empirically, b has often been found to be close to 1/3, which would be the power exponent of R when the equation is expressed in its original (non-log) form. However, empirically the exponent can and often does depart from 1/3.

Rick notes that this regression analysis omits log D as a predictor variable. He asserts that it is this omission that allows the exponent to depart from 1/3, and that if log D is included, then the mathematics of the calculation of the radius of curvature necessarily impose a 1/3 exponent in the formula relating V to R.

In reaching this conclusion, Rick has fallen into a mathematical fallacy, because the radius of curvature itself depends on D. It is the division of V3 by D that makes R a pure measure of distance, uninfluenced by the velocity of the point that moves along the curve.

To better understand this assertion, consider my parallel construction “proving” that the velocity of a point traversing a circle depends on the circle’s area. Obviously this is nonsense – you can go around the circle at any velocity whatsoever no matter what the area of the circle may be. Yet my equation seems to indicate that the velocity of the point is mathematically determined by the circle’s area. How can this be? Well, let’s have a look at my original equation again:

V = 2A1/2π1/2/T

The key is to understand the role of T. Let’s move T to the other side of the equation:

VT = 2A1/2π1/2

V is the velocity of the point in, say, cm/second. T is the time in seconds required to traverse the circle at this particular velocity

V cm/s times T s = cm, a distance. In fact it is the distance around the circle ordinarily referred to as the circle’s circumference. I’ll call it C.

So the formula can now be rewritten as

C = 2A1/2π1/2.

I think you will agree that the area of a circle depends on its circumference. The two measured variables, V and T, were used as a means of determining the circumference. The faster you go around the circle, the sooner you will complete the circuit, so the two variables vary inversely in such a way that the circumference that you determine from V and T will be the same for a given circle no matter what speed you go around it at.

By exactly the same logic, in the formula for the radius of curvature, the denominator D of the formula compensates for the different velocities at which the point may be moving along the curve, so that when V3 is divided by D, the result, the radius, is purely a measure of distance that will be the same value no mater what the velocity of the point was observed to be when the curve was drawn. Because R as computed is independent of the velocity used to compute it, the empirically observed relationship between V and R could be any relationship whatever, and the empirically observed power-law relationship is not an artifact of the way in which R is computed.

It’s as simple as that.

Bruce

[From Bruce Abbott (2018.01.01.1145 EST)]

I have discovered a heretofore unsuspected relationship between the area of a circle and the speed with which the circumference of the circle is traced.

A point traces the circumference of a circle at velocity V, completing the transit in time T. Then the area A of the circle is given by

A = V2T2/4π

Solving for V:

V = 2A1/2π1/2/T

Notice that the velocity with which the circle is traced is a power-law function of the circle’s area, with an exponent of 1/2. Converting to logarithms:

Log V = log 2 + ½ log A + ½ log π – log T

Where it appears that velocity of the circumscribing point and the area of the circle are independent, this is only due to the influence of log T, which allows the relationship to depart from what otherwise would be a mathematically enforced power relationship.

It amazes me that professional mathematicians have overlooked this important fact!

Bruce

bruce, why did you spend so much of your time trying to correct bullshit in so many different lengthy ways?

···

Alex Gomez-Marin, PhD

Instituto de Neurociencias

behavior-of-organisms.org

[Martin Taylor 2018.04.03.07.38]

``````  Alex, don't you have an "April Fool's Day" tradition? Some of us
``````

enjoy crafting a spoof message for April 1 from time to time. It’s
even more fun if there’s a point behind the spoof. I don’t know a
PCT explanation of “to have fun”, but for some, I suspect most,
people, to have a bit of fun is a goal whether there is anything
behind it or not.

Martin

···

On 2018/04/3 2:03 AM, Alex Gomez-Marin
wrote:

``````      bruce, why did you spend so much of your time
``````

trying to correct bullshit in so many different lengthy ways?

On Mon, 2 Apr 2018 at 23:35, Bruce Abbott <bbabbott@frontier.com >
wrote:

``````              [From Bruce Abbott (2018.04.02.1735
``````

EST)]

Â

``````              Correction: The superscript code
``````

somehow got lost in the formula for R: it reads R =
V3/D but should read V3 /D.Â I have
corrected it in the text below.

Â

Bruce

Â

``````              [From
``````

Bruce Abbott (2018.04.02.1725 EST)]

Â

``````              For those
``````

of you who may have been taken in by my post
(reproduced at bottom below), April Fool!

Â

``````              But there
``````

was a serious purpose in posting it.Â I had presented
the following equation:

Â

V = 2A1/2Ï€1/2/T

Â

``````              This form
``````

parallels Rick Markenâ€™s analysis (Marken &
Shaffer, 2017) of the formula for computing the radius
of curvature at a given point as the point traces some
curved path.Â That formula is as follows:

Â

R = V3 /D,
where V is the velocity of the point along the curve
and D is computed from the velocities and
accelerations of the point along the X and Y
directions.Â Solving for V yields:

Â

V = R1/3/D1/3 ,
which is similar in form to my equation above relating
velocity to circle area.

Â

``````              Converted
``````

to log form, my equation becomes:

Â

``````              log V =
``````

log 2 + Â½ log A + Â½ log Ï€ – log T<

Â

``````              And Rickâ€™s
``````

becomes:

Â

``````              log V =
``````

1/3 log R - 1/3 log D

Â

``````              Empirically,
``````

a power-law relationship is found between V and R.Â
This is obtained by doing a linear regression analysis
in which log R is the predictor variable and log V is
the criterion variable:

Â

``````              log V = a
``````
• b*log R,

Â

``````              where a is
``````

the intercept of the best fitting line relating log V
to log R and b is the slope.Â Empirically, b has often
been found to be close to 1/3, which would be the
power exponent of R when the equation is expressed in
its original (non-log) form.Â However, empirically the
exponent can Â and often does depart from 1/3.

Â

``````              Rick notes
``````

that this regression analysis omits log D as a
predictor variable.Â He asserts that it is this
omission that allows the exponent to depart from 1/3,
and that if log D is included, then the mathematics of
the calculation of the radius of curvature necessarily
impose a 1/3 exponent in the formula relating V to R.

Â

``````              In
``````

reaching this conclusion, Rick has fallen into a
mathematical fallacy, because the radius of curvature
itself depends on D.Â It is the division of V3
by D that makes Â R a pure measure of distance,
uninfluenced by the velocity of the point that moves
along the curve.

Â

``````              To better
``````

understand this assertion, consider my parallel
construction â€œprovingâ€? that the velocity of a point
traversing a circle depends on the circleâ€™s area.Â
Obviously this is nonsense – you can go around the
circle at any velocity whatsoever no matter what the
area of the circle may be.Â Yet my equation seems to
indicate that the velocity of the point is
mathematically determined by the circleâ€™s area.Â How
can this be?Â Well, letâ€™s have a look at my original
equation again:

Â

V = 2A1/2Ï€1/2/T

Â

``````              The key is
``````

to understand the role of T.Â Letâ€™s move T to the
other side of the equation:

Â

VT = 2A1/2Ï€1/2

Â

``````              V is the
``````

velocity of the point in, say, cm/second.Â T is the
time in seconds required to traverse the circle at
this particular velocity

Â

``````              V cm/s
``````

times T s = cm, a distance.Â In fact it is the
distance around the circle ordinarily referred to as
the circleâ€™s circumference.Â Iâ€™ll call it C.

Â

``````              So the
``````

formula can now be rewritten as

Â

C = 2A1/2Ï€ 1/2.

Â

``````              I think
``````

you will agree that the area of a circle depends on
its circumference.Â The two measured variables, V and
T, were used as a means of determining the
circumference.Â The faster you go around the circle,
the sooner you will complete the circuit, so the two
variables vary inversely in such a way that the
circumference that you determine from V and T will be
the same for a given circle no matter what speed you
go around it at.

Â

``````              By exactly
``````

the same logic, in the formula for the radius of
curvature, the denominator D of the formula
compensates for the different velocities at which the
point may be moving along the curve, so that when V3
is divided by D, the result, the radius, is purely a
measure of distance that will be the same value no
mater what the velocity of the point was observed to
be when the curve was drawn.Â Because R as computed is
independent of the velocity used to compute it, the
empirically observed relationship between V
and R could be any relationship whatever, and the
empirically observed power-law relationship is not an
artifact of the way in which R is computed.

Â

``````              Itâ€™s as
``````

simple as that.

Â

Bruce

Â

``````              [From
``````

Bruce Abbott (2018.01.01.1145 EST)]

Â

``````              I have
``````

discovered a heretofore unsuspected relationship
between the area of a circle and the speed with which
the circumference of the circle is traced.

Â

``````              A point
``````

traces the circumference of a circle at velocity V,
completing the transit in time T.Â Then the area A of
the circle is given by

Â

A = V2T2/4Ï€

Â

``````              Solving
``````

for V:

V = 2A1/2Ï€1/2/T

Â

``````              Notice
``````

that the velocity with which the circle is traced is a
power-law function of the circleâ€™s area, with an
exponent of 1/2.Â Converting to logarithms:

Â

``````              Log V =
``````

log 2 + Â½ log A + Â½ log Ï€ – log T<

Â

``````              Where it
``````

appears that velocity of the circumscribing
point and the area of the circle are independent, this
is only due to the influence of log T, which allows
the relationship to depart from what otherwise would
be a mathematically enforced power relationship.

Â

``````              It amazes
``````

me that professional mathematicians have overlooked
this important fact!

Â

Bruce

Â

Â

Alex Gomez-Marin, PhD

Instituto de Neurociencias

behavior-of-organisms.org

Sure, sure. I like having fun too. And our “fool’s day” in Spain is the 28th of December. What I found funny and worrisome at the same time is that I seriously thought that such joke was serious given what often goes on at CSGnet…

···

On Tue, Apr 3, 2018 at 1:43 PM, Martin Taylor mmt-csg@mmtaylor.net wrote:

[Martin Taylor 2018.04.03.07.38]

``````  Alex, don't you have an "April Fool's Day" tradition? Some of us
``````

enjoy crafting a spoof message for April 1 from time to time. It’s
even more fun if there’s a point behind the spoof. I don’t know a
PCT explanation of “to have fun”, but for some, I suspect most,
people, to have a bit of fun is a goal whether there is anything
behind it or not.

Martin

``````  On 2018/04/3 2:03 AM, Alex Gomez-Marin
``````

wrote:

``````      bruce, why did you spend so much of your time
``````

trying to correct bullshit in so many different lengthy ways?

On Mon, 2 Apr 2018 at 23:35, Bruce Abbott <bbabbott@frontier.com >
wrote:

``````              [From Bruce Abbott (2018.04.02.1735
``````

EST)]

Â

``````              Correction: The superscript code
``````

somehow got lost in the formula for R: it reads R =
V3/D but should read V3 /D.Â I have
corrected it in the text below.

Â

Bruce

Â

``````              [From
``````

Bruce Abbott (2018.04.02.1725 EST)]

Â

``````              For those
``````

of you who may have been taken in by my post
(reproduced at bottom below), April Fool!

Â

``````              But there
``````

was a serious purpose in posting it.Â I had presented
the following equation:

Â

V = 2A1/2Ï€1/2/T

Â

``````              This form
``````

parallels Rick Markenâ€™s analysis (Marken &
Shaffer, 2017) of the formula for computing the radius
of curvature at a given point as the point traces some
curved path.Â That formula is as follows:

Â

R = V3 /D,
where V is the velocity of the point along the curve
and D is computed from the velocities and
accelerations of the point along the X and Y
directions.Â Solving for V yields:

Â

V = R1/3/D1/3 ,
which is similar in form to my equation above relating
velocity to circle area.

Â

``````              Converted
``````

to log form, my equation becomes:

Â

``````              log V =
``````

log 2 + Â½ log A + Â½ log Ï€ – log T<

Â

``````              And Rickâ€™s
``````

becomes:

Â

``````              log V =
``````

1/3 log R - 1/3 log D

Â

``````              Empirically,
``````

a power-law relationship is found between V and R.Â
This is obtained by doing a linear regression analysis
in which log R is the predictor variable and log V is
the criterion variable:

Â

``````              log V = a
``````
• b*log R,

Â

``````              where a is
``````

the intercept of the best fitting line relating log V
to log R and b is the slope.Â Empirically, b has often
been found to be close to 1/3, which would be the
power exponent of R when the equation is expressed in
its original (non-log) form.Â However, empirically the
exponent can Â and often does depart from 1/3.

Â

``````              Rick notes
``````

that this regression analysis omits log D as a
predictor variable.Â He asserts that it is this
omission that allows the exponent to depart from 1/3,
and that if log D is included, then the mathematics of
the calculation of the radius of curvature necessarily
impose a 1/3 exponent in the formula relating V to R.

Â

``````              In
``````

reaching this conclusion, Rick has fallen into a
mathematical fallacy, because the radius of curvature
itself depends on D.Â It is the division of V3
by D that makes Â R a pure measure of distance,
uninfluenced by the velocity of the point that moves
along the curve.

Â

``````              To better
``````

understand this assertion, consider my parallel
construction â€œprovingâ€? that the velocity of a point
traversing a circle depends on the circleâ€™s area.Â
Obviously this is nonsense – you can go around the
circle at any velocity whatsoever no matter what the
area of the circle may be.Â Yet my equation seems to
indicate that the velocity of the point is
mathematically determined by the circleâ€™s area.Â How
can this be?Â Well, letâ€™s have a look at my original
equation again:

Â

V = 2A1/2Ï€1/2/T

Â

``````              The key is
``````

to understand the role of T.Â Letâ€™s move T to the
other side of the equation:

Â

VT = 2A1/2Ï€1/2

Â

``````              V is the
``````

velocity of the point in, say, cm/second.Â T is the
time in seconds required to traverse the circle at
this particular velocity

Â

``````              V cm/s
``````

times T s = cm, a distance.Â In fact it is the
distance around the circle ordinarily referred to as
the circleâ€™s circumference.Â Iâ€™ll call it C.

Â

``````              So the
``````

formula can now be rewritten as

Â

C = 2A1/2Ï€ 1/2.

Â

``````              I think
``````

you will agree that the area of a circle depends on
its circumference.Â The two measured variables, V and
T, were used as a means of determining the
circumference.Â The faster you go around the circle,
the sooner you will complete the circuit, so the two
variables vary inversely in such a way that the
circumference that you determine from V and T will be
the same for a given circle no matter what speed you
go around it at.

Â

``````              By exactly
``````

the same logic, in the formula for the radius of
curvature, the denominator D of the formula
compensates for the different velocities at which the
point may be moving along the curve, so that when V3
is divided by D, the result, the radius, is purely a
measure of distance that will be the same value no
mater what the velocity of the point was observed to
be when the curve was drawn.Â Because R as computed is
independent of the velocity used to compute it, the
empirically observed relationship between V
and R could be any relationship whatever, and the
empirically observed power-law relationship is not an
artifact of the way in which R is computed.

Â

``````              Itâ€™s as
``````

simple as that.

Â

Bruce

Â

``````              [From
``````

Bruce Abbott (2018.01.01.1145 EST)]

Â

``````              I have
``````

discovered a heretofore unsuspected relationship
between the area of a circle and the speed with which
the circumference of the circle is traced.

Â

``````              A point
``````

traces the circumference of a circle at velocity V,
completing the transit in time T.Â Then the area A of
the circle is given by

Â

A = V2T2/4Ï€

Â

``````              Solving
``````

for V:

V = 2A1/2Ï€1/2/T

Â

``````              Notice
``````

that the velocity with which the circle is traced is a
power-law function of the circleâ€™s area, with an
exponent of 1/2.Â Converting to logarithms:

Â

``````              Log V =
``````

log 2 + Â½ log A + Â½ log Ï€ – log T<

Â

``````              Where it
``````

appears that velocity of the circumscribing
point and the area of the circle are independent, this
is only due to the influence of log T, which allows
the relationship to depart from what otherwise would
be a mathematically enforced power relationship.

Â

``````              It amazes
``````

me that professional mathematicians have overlooked
this important fact!

Â

Bruce

Â

Â

Alex Gomez-Marin, PhD

Instituto de Neurociencias

behavior-of-organisms.org

Alex Gomez-Marin, PhD

Instituto de Neurociencias

behavior-of-organisms.org

[From Erling Jorgensen (2018.04.04 1020 EST)]

Bruce Abbott (2018.04.02.1725 EST) & (2018.04.02.1735 EST)

For those of you who may have been taken in by my post (reproduced at bottom below), April Fool!

But there was a serious purpose in posting it. …

Hi Bruce,

EJ: It’s the serious purpose that I want to pick up on. I found you ‘parable’ and explanations extremely helpful in understanding what is going on ‘inside’ the power-law equations, as it were.

EJ: As I’ve admitted at other times, math is not a main avenue of expression for me; but a doctoral degree is at least good for following the gist of a mathematical argument. I have known that something did not feel right about the Marken & Shafer (2017) treatment of the power-law issue. I did try to convey that in a few posts at the height of the debate last Fall on CSGnet.

EJ: I have known that an equation with velocity on both sides of the equals sign has not been sufficiently reduced or factored properly. I have known that you cannot use velocity as a predictor of itself. That is just stating a tautology. I have known that whenever an equation’s prediction exactly matches observed results, that should make us question the set-up further. And it is further evidence that a tautology may be at work. I have also had a suspicion (but this may be my non-math bias) that any factor added to an equation ought to have some grounding or rationale, explainable in words, beyond the mathematical manipulation itself. I have yet to hear that about whatever “affine velocity” is supposed to represent.

EJ: Beyond that, I have had to lean on those I trust who were more adept at math, in sorting through these arguments. And that has included you, Martin T., and Richard K.

EJ: It wasn’t until your Parable of the Area of a Circle, if I may call it such, that the manipulations behind the Gribble & Ostry (1996) equation became clear. Your explication here has been very helpful, as well as convincing.

EJ: It makes sense that traveling at a certain speed for a certain period of time yields a given distance. It makes sense that traveling at a faster speed for a shorter period of time can yield the same distance. It thus makes sense that such an operation can stand in for the distance measure itself, where time gets canceled out of the factor. That for me has been a useful way to understand the Gribble & Ostry (1996) manipulations. Until now, I haven’t known why they were multiplying speed in an X direction by acceleration in a Y direction, as well as speed in the Y direction by acceleration in the X direction. Your explanation takes the ‘mumbo-jumbo’ out of that for me. Looking back at posts I have copied on this topic, I realize you explained this before in [Bruce Abbott (2017.11.1640)] – I think the full date should have been 2017.11.12 … – but the current Parable seems to work better for my brain.

EJ: I still don’t have much intuitive sense of why a cube-root of some kind keeps showing up in these studies. My mental place-holder for now is that maybe it has something to do with necessary changes in a second derivative, i.e., acceleration, whether that is happening by a person’s hand or a larva’s body.

All the best,

Erling

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[Martin Taylor 2018.04.04.12.56]

``````Yes, Initially I had the same problem, but after a little while I
``````

realized that what you say here is on the right track. Here’s one
way to look at it.
The denominator expression is written in the various papers
(x-doty-doubledot - y-dotxdoubledot), I presume to save space on
the page. In the ordinary notation, that is expressed as (dx/dt*dy/dt

• dy/dtdx/dt). The "dt"s in the
denominators are not mysterious. They are all tiny stretches of
time, and in the calculations from sampled data, they are the time
between successive samples. So we can rewrite that denominator
expressions as (dx
dy-dy*dx)/dt.
The cubed inter-sample time is where the cube roots come from. But before we can be sure of this, we have to see whether there is
any compensating effect in the numerator, because it also has
superscript "2"s that look the same as the "2"s that indicate
“square-me”. But they aren’t. The two kinds of superscript “2” are
like homophones in a language, “be” and “bee”, for example, that
sound the same but mean quite different things. “dx” and “dy” are
the distance the moving thing moved in one sample interval. “dx”
is the difference between the distance “dx” measured for one pair of
consecutive samples and “dx” over the next pair. The same goes, of
course, for dy. So there is no compensating effect, and if you want to think of the
variables in any following equations to be moving through time, you
want to turn “dt” into "dt’ by taking the cube root of
everything. You don’t have to, but if you want to have an expression
for velocity, that is simply ds/dt, where “s” is a distance
travelled in time dt. If what you have at hand is dt-cubed, you will
be calculating velocity-cubed. In this case that is not what you
want. Mathematics is just a language, and it’s nobody’s native language.
If you don’t use it much, you tend to lose it (as have I to a large
extent). If you have had a year or two of French at school, you
probably can enquire about hotel rooms and train time-tables or
in French (or English) questions about the subtleties of his
thought, or appreciate the jokes in a Molière play. It’s the same
with Math, but whether in French or Math, sometimes things do mean
what they seem to mean. The "dx"s and "dy"s in the denominators of
differential expressions are such things.
Martin
···

On 2018/04/4 11:15 AM, Erling Jorgensen
wrote:

[From Erling Jorgensen (2018.04.04 1020 EST)]

``````      >Bruce Abbott (2018.04.02.1725 EST)   &
``````

(2018.04.02.1735 EST)

``````      >For those of you who may have been taken in by my post
``````

(reproduced at bottom below), April Fool!

``````      >But there was a serious purpose in
``````

posting it. …

Hi Bruce,

``````      EJ:  It's the serious purpose that I want to pick up on.  I
``````

found you ‘parable’ and explanations extremely helpful in
understanding what is going on ‘inside’ the power-law
equations, as it were.

``````      EJ:  I still don't have much intuitive sense of why a
``````

cube-root of some kind keeps showing up in these studies. My
mental place-holder for now is that maybe it has something to
do with necessary changes in a second derivative, i.e.,
acceleration, whether that is happening by a person’s hand or
a larva’s body.

2222223

2

3

[From Bruce Abbott (2018.04.04.1950 EST)]

[From Erling Jorgensen (2018.04.04 1020 EST)]

Bruce Abbott (2018.04.02.1725 EST) & (2018.04.02.1735 EST)

For those of you who may have been taken in by my post (reproduced at bottom below), April Fool!

But there was a serious purpose in posting it. …

Hi Bruce,

EJ: It’s the serious purpose that I want to pick up on. I found you ‘parable’ and explanations extremely helpful in understanding what is going on ‘inside’ the power-law equations, as it were.

EJ: As I’ve admitted at other times, math is not a main avenue of expression for me; but a doctoral degree is at least good for following the gist of a mathematical argument. I have known that something did not feel right about the Marken & Shafer (2017) treatment of the power-law issue. I did try to convey that in a few posts at the height of the debate last Fall on CSGnet.

EJ: I have known that an equation with velocity on both sides of the equals sign has not been sufficiently reduced or factored properly. I have known that you cannot use velocity as a predictor of itself. That is just stating a tautology. I have known that whenever an equation’s prediction exactly matches observed results, that should make us question the set-up further. And it is further evidence that a tautology may be at work. I have also had a suspicion (but this may be my non-math bias) that any factor added to an equation ought to have some grounding or rationale, explainable in words, beyond the mathematical manipulation itself. I have yet to hear that about whatever “affine velocity” is supposed to represent.

EJ: Beyond that, I have had to lean on those I trust who were more adept at math, in sorting through these arguments. And that has included you, Martin T, and Richard K.

EJ: It wasn’t until your Parable of the Area of a Circle, if I may call it such, that the manipulations behind the Gribble & Ostry (1996) equation became clear. Your explication here has been very helpful, as well as convincing.

EJ: It makes sense that traveling at a certain speed for a certain period of time yields a given distance. It makes sense that traveling at a faster speed for a shorter period of time can yield the same distance. It thus makes sense that such an operation can stand in for the distance measure itself, where time gets canceled out of the factor. That for me has been a useful way to understand the Gribble & Ostry (1996) manipulations. Until now, I haven’t known why they were multiplying speed in an X direction by acceleration in a Y direction, as well as speed in the Y direction by acceleration in the X direction. Your explanation takes the ‘mumbo-jumbo’ out of that for me. Looking back at posts I have copied on this topic, I realize you explained this before in [Bruce Abbott (2017.11.1640)] – I think the full date should have been 2017.11.12 … – but the current Parable seems to work better for my brain.

EJ: I still don’t have much intuitive sense of why a cube-root of some kind keeps showing up in these studies. My mental place-holder for now is that maybe it has something to do with necessary changes in a second derivative, i.e., acceleration, whether that is happening by a person’s hand or a larva’s body.

Thanks, Erling, I appreciate the feedback!

It may be a surprise to you, but my own math abilities are probably no better than yours and may be worse. However, when this debate first surfaced, I searched the Internet for information on the computation of curvature and found some very helpful sources. One of these is a series of short tutorials on curvature presented as YouTube lectures by Sal Kahn of the Kahn Academy. The first of these gives you a geometrical feel for what the mathematical concept of curvature is. This is followed by additional tutorial videos that help one to understand how the formula for computing curvature (or its inverse, the radius of curvature) arises mathematically. The essential concept is that a curve of varying curvature can be fit at any specific point by a best-fitting circle of a given radius, called the osculating circle. The radius of this circle is the radius of curvature, and its inverse is the curvature.

Curvature thus applies to a given point along a curve that may be varying in the tightness of its curves, and thus may vary from point to point. To determine the radius of the curve at a given point, one could have the point moving with a specific velocity. As it approaches the spot where we want to know the curvature, the point is not only moving along the curve, it is changing direction along both the X and Y axes. How fast it is changing direction depends on the curvature (and on the velocity). By using a little math, one can find the rates of change at the specific point along the curve for which the curvature is to be obtained. Tracing the curve at a higher velocity changes not only the velocity of the point along the curve, but also the accelerations and velocities in the X and Y directions as the path curves. The formula for radius of curvature is such that the same curvature results no mater what the velocity of the point at that location on the curve.

In the studies in which a curved path is being traced at some velocity, the X,Y position of the point is recorded as a function of time. From the differences in position from one sample to the next, one can estimate the point’s velocity along the curve as well as its velocities and accelerations along the X and Y directions; these estimates are then plugged into the formulas presented by Gribble and Ostry to yield estimates of both the velocity of the point along the curve and the radius of curvature, at each sampled interval.

If you are interested enough to investigate further, I recommend those Khan Academy videos. They are relatively short (about 5 minutes each, I think) and astonishingly easy to follow. The first one is located at