Gene Fuller wrote:
How do you know that "reality" is correct and the "math model" is wrong?

With a ridiculous question like that, I rest my case! Reality is
*always* correct. If the math model disagrees with reality, it is
simply wrong! Do you really think that what happens only in your
mind is reality? As my Mother once said, "If that's what you think,
think again!"

Example: The 19th century math model didn't explain the orbit of
Mercury. Do you think you can control the orbit of Mercury in
your mind? I suppose in your world, you really can do that.

The second paragraph is even more curious. Do you have a reference for
this migration of energy from the H-field to supply the suffering
E-field? I must have missed that day in class.

You were probably out with a cold that day in kindergarten. It's
called "conservation of energy". If the ratio of the E-field to
the H-field is a constant, any change in either one will result
in a change in the other. This is a well known and accepted fact
of physics in the field of optics. How did you miss it? Hint: What
happens to isotropic radiation in 3D space? Both the E-field and
H-field decrease in value per square meter and THEIR RATIO STAYS
EXACTLY THE SAME.

FYI, the characteristic impedance or index of refraction forces the
ratio of the E-field to the H-field to a constant value. When a
forward wave in a 50 ohm transmission line encounters a 50 ohm
to 300 ohm impedance discontinuity, the ratio of the voltage to
current in the forward wave changes from 50 to 300. I'm extremely
surprised that you don't know and don't accept that fact of physics.
--
73, Cecil, W5DXP

On Fri, 19 Nov 2004 17:59:01 -0600, Cecil Moore
wrote:
I suppose in your world, you really can do that.
21st Century math doesn't explain the orbit of pluto, but 28th century
math will....

Richard Clark wrote:
On Fri, 19 Nov 2004 17:59:01 -0600, Cecil Moore
wrote:

I suppose in your world, you really can do that.

21st Century math doesn't explain the orbit of pluto, but 28th century
math will....

We have all the math we need to handle Pluto's orbit, and have had for a
century. We do not, however, know all the bits out there that may
perturb it. That said, we can still predict it quite well enough for
anyone's current needs.

tom
K0TAR

On Fri, 19 Nov 2004 21:04:45 -0600, Tom Ring
wrote:
That said, we can still predict it quite well enough for
anyone's current needs.

Hi Tom,

Exactly, but you were responding to a dry comment on the nature of
"reality vs. models" which was absurd from the get-go.

To say that reality drives models or vice-versa (especially from
armchair theorists) is naive in the extreme. Anyone's "knowledge" of
reality is simply its own corrupt model and not any particular state
of enlightenment. The naive part is in not knowing the errors
bounding this "knowledge."

Barring revolution, anarchy, social upset or the rest, futurity will
tend generally to more resolution in the answer and a finer reduction
of error, but it will never be fully resolved because the answer could
not be encompassed by the mind.

73's
Richard Clark, KB7QHC

Richard Clark wrote:

On Fri, 19 Nov 2004 21:04:45 -0600, Tom Ring
wrote:

That said, we can still predict it quite well enough for
anyone's current needs.


Hi Tom,

Exactly, but you were responding to a dry comment on the nature of
"reality vs. models" which was absurd from the get-go.

To say that reality drives models or vice-versa (especially from
armchair theorists) is naive in the extreme. Anyone's "knowledge" of
reality is simply its own corrupt model and not any particular state
of enlightenment. The naive part is in not knowing the errors
bounding this "knowledge."

Barring revolution, anarchy, social upset or the rest, futurity will
tend generally to more resolution in the answer and a finer reduction
of error, but it will never be fully resolved because the answer could
not be encompassed by the mind.

73's
Richard Clark, KB7QHC

Including revolution, anarchy, social upset or the rest, the minds will
encompass everything to the finest resolution, because they will be the
ones we built to surpass us. Unless we destroy ourselves first, it is
inevitable.

tom
K0TAR

Cecil,

You nicely ducked the question, again.

This is not about philosophy, existentialism, the meaning of life,
planetary orbits, or any other such fluff.

Again, how do you know the "reality" for your transmission line in free
space?

This is a straightforward question. Do you have data? Do you have a
reference that you consider reality? Are you basing your reality on a
model? Is it just a matter of blind faith? Something else?

You have many times made similar remarks about us "stupid folks" who get
misled by math models. How do you get your information?


As for the E- and H-fields, this just gets more amusing by the minute.

73,
Gene
W4SZ

Cecil Moore wrote:
Gene Fuller wrote:

How do you know that "reality" is correct and the "math model" is wrong?


With a ridiculous question like that, I rest my case! Reality is
*always* correct. If the math model disagrees with reality, it is
simply wrong! Do you really think that what happens only in your
mind is reality? As my Mother once said, "If that's what you think,
think again!"

Example: The 19th century math model didn't explain the orbit of
Mercury. Do you think you can control the orbit of Mercury in
your mind? I suppose in your world, you really can do that.

The second paragraph is even more curious. Do you have a reference for
this migration of energy from the H-field to supply the suffering
E-field? I must have missed that day in class.


You were probably out with a cold that day in kindergarten. It's
called "conservation of energy". If the ratio of the E-field to
the H-field is a constant, any change in either one will result
in a change in the other. This is a well known and accepted fact
of physics in the field of optics. How did you miss it? Hint: What
happens to isotropic radiation in 3D space? Both the E-field and
H-field decrease in value per square meter and THEIR RATIO STAYS
EXACTLY THE SAME.

FYI, the characteristic impedance or index of refraction forces the
ratio of the E-field to the H-field to a constant value. When a
forward wave in a 50 ohm transmission line encounters a 50 ohm
to 300 ohm impedance discontinuity, the ratio of the voltage to
current in the forward wave changes from 50 to 300. I'm extremely
surprised that you don't know and don't accept that fact of physics.
--
73, Cecil, W5DXP

Gene Fuller wrote:
Again, how do you know the "reality" for your transmission line in free
space?

Simple, we know the characteristics of copper wire and we know the
dielectric properties of a vacuum. Examples abound in textbooks.

This is a straightforward question. Do you have data?

Of course, don't you? If not, maybe you had better get some.

You have many times made similar remarks about us "stupid folks" who get
misled by math models. How do you get your information?

The distributed network model was developed because the circuit model
failed in systems that are an appreciable portion of a wavelength. I'm
surprised that you don't have that information.

As for the E- and H-fields, this just gets more amusing by the minute.

The attenuation factor for the average transmission line at HF is almost
entirely due to the resistance in the conductors as the shunt conductance
is usually negligible. Resistance in the conductors causes a voltage drop.
Yet, we know that the current is attenuated by exactly the same percentage
as the voltage. Since G is negligible, R must be responsible for the
decrease in current. What laws of physics can account for that fact?

Z0 determines the ratio of HF voltage to HF current. Therefore, if the
voltage drops and the V/I ratio stays constant, energy must be transferred
from the H-field to the E-field. That's pretty simple stuff, Gene, maybe
sophomore level.
--
73, Cecil, W5DXP

Gene Fuller wrote:
As for the E- and H-fields, this just gets more amusing by the minute.
The second paragraph is even more curious. Do you have a reference
for this migration of energy from the H-field to supply the suffering
E-field?

Here's the reference: From "Fields and Waves" by Ramo and Whinnery.
Take a close look at the exponential transmission line equations
for flat lines (no reflections):

V = Vmax(e^-az)(e^wt-bz) (1)

I = Vmax(e^-az)(e^wt-bz)/Z0 (2)

'a' (alpha) is the attenuation factor. The two equations are identical
except for the Z0 term. If you divide equation (1) by equation (2),
you get Z0. In a flat transmission line (no reflections) the current
is ALWAYS equal to the voltage divided by the characteristic impedance
of the transmission line. The voltage and current are attenuated by
EXACTLY the same factor. If the voltage drops because of I^2*R losses,
the current must decrease by exactly the same percentage. (I have avoided
calling it a current drop so it wouldn't upset you.)

Since the attenuation factor is R/2*Z0 + G*Z0/2 and since, for most
transmission lines used on HF, R/2*Z0 G*Z0/2, the current attenuation
is caused by the series I^2*R drop in the voltage and the V/I=Z0 ratio
that must be maintained - pretty simple logic.

I must have missed that day in class.

Yep, you must have. But it's not too late to learn what you missed
that day. :-)
--
73, Cecil, W5DXP

Cecil,

Nice try.

I cannot find the word "migrate" in your reference. However, I do find
several examples of your own interpretations mixed in.

When in doubt, change the subject?

More rattlesnake physics?

73,
Gene
W4SZ

Cecil Moore wrote:
Gene Fuller wrote:

As for the E- and H-fields, this just gets more amusing by the minute.

The second paragraph is even more curious. Do you have a reference
for this migration of energy from the H-field to supply the
suffering E-field?


Here's the reference: From "Fields and Waves" by Ramo and Whinnery.
Take a close look at the exponential transmission line equations
for flat lines (no reflections):

V = Vmax(e^-az)(e^wt-bz) (1)

I = Vmax(e^-az)(e^wt-bz)/Z0 (2)

'a' (alpha) is the attenuation factor. The two equations are identical
except for the Z0 term. If you divide equation (1) by equation (2),
you get Z0. In a flat transmission line (no reflections) the current
is ALWAYS equal to the voltage divided by the characteristic impedance
of the transmission line. The voltage and current are attenuated by
EXACTLY the same factor. If the voltage drops because of I^2*R losses,
the current must decrease by exactly the same percentage. (I have avoided
calling it a current drop so it wouldn't upset you.)

Since the attenuation factor is R/2*Z0 + G*Z0/2 and since, for most
transmission lines used on HF, R/2*Z0 G*Z0/2, the current attenuation
is caused by the series I^2*R drop in the voltage and the V/I=Z0 ratio
that must be maintained - pretty simple logic.

I must have missed that day in class.


Yep, you must have. But it's not too late to learn what you missed
that day. :-)
--
73, Cecil, W5DXP

Gene Fuller wrote:
I cannot find the word "migrate" in your reference. However, I do find
several examples of your own interpretations mixed in.

You expect me to remember the exact wording after 40 years?
Perhaps your prof, like mine, told you to skip sections 1.22-
1.28 in Ramo and Whinnery. I didn't take his advice - I read
them.

So Gene, please point out the error in my logic. If the
current attenuation is primarily from R, the series
resistance, what other explanation can there be than
energy is being supplied by the H-field to the sagging
E-field? Can you think of any other rational, logical,
non-emotional, technical explanation, given these
exponential transmission line equations?

V = Vmax(e^-az)*e^j(wt-bz)

I = Vmax(e^-az)*e^j(wt-bz)/Z0

V/I = Z0

(I think you are just trying to punish me for saying my
dog has a soul. :-)
--
73, Cecil, W5DXP