Cecil Moore wrote:
I have *NEVER* said "power is moving", at least not in this
century.

Do you imagine that the caviat "not in this century" might make the
statement at least partially true?

That is just your straw man raising its ugly head
yet once again.

Was it not raised when you brought the definition of 'transfer' back
into the discussion - again this century?

ac6xg

Jim Kelley wrote:
Cecil Moore wrote:
I have *NEVER* said "power is moving", at least not in this
century.

Do you imagine that the caviat "not in this century" might make the
statement at least partially true?

In the 20th century, I did believe in power flow but you
convinced me that I was wrong and I changed my mind. I
have not believed in power flow during the 21st century.


That is just your straw man raising its ugly head
yet once again.

Was it not raised when you brought the definition of 'transfer' back
into the discussion - again this century?

I will keep bringing it up until you furnish the definition
that you are using for the word. That you absolutely refuse
to provide a definition means it is nothing but your gut
feeling about the matter.
--
73, Cecil http://www.w5dxp.com

Richard Clark wrote:
On Thu, 13 Dec 2007 08:40:53 -0800, Roger wrote:


And just for completeness...
The fundamental equations also work when:
- the signal is not sinusoidal, e.g. pulse, step, square, ...
- rather than a load at one end, there is a source at each end
- the sources at each end produce different arbitrary functions
- the arbitrary functions at each end are DC sources
It is highly instructive to compute the forward and reverse
voltage and current (and then power) for a line with the same
DC voltage applied to each end.

...Keith

...Keith


Interesting! The important thing is to get answers that agree with
our experiments.

I have done some computations for DC voltage applied to transmission
lines. The real surprise for me came when I realized that transmission
line impedance could be expressed as a function of capacitance and the
wave velocity. Z0 = 1/cC where c is the velocity of the wave and C is
the capacitance of the transmission line per unit length.


Hi Roger,

This last round has piqued my interest when we dipped into DC. Those
"formulas" would lead us to a DC wave velocity?

73's
Richard Clark, KB7QHC

Hi Richard,

Here are two links to pages that cover the derivation of the formula Zo
= 1/cC and much more.

http://www.speedingedge.com/PDF-File..._Impedance.pdf
http://www.ece.uci.edu/docs/hspice/h...001_2-269.html

Here is the way I proposed to Kevin Schmidt nearly seven years ago after
seeing him use the formula on a web page:

*ASSUME*:
1) An electrical wave travels at the speed of light, c
2) A 'perfect' voltage source without impedance, V
3) A 'perfect' transmission line having no resistance but uniform
capacitance per unit length, C

*CONDITIONS AND SOLUTION*
The perfect voltage source has one terminal connected to the
transmission line prior to beginning the experiment. The experiment
begins by connecting the second terminal to the transmission line. The
voltage source drives an electrical wave down the transmission line at
the speed of light. Because of the limitation of speed, the wave
travels in the shape of a square wave containing all frequencies
required to create a square wave.

The square wave travels down the transmission line at the speed of light
(c). After time (T), the wave has traveled distance cT down the
transmission line, and has charged the distributed capacity CcT of the
line to voltage V over that distance. The total charge Q on the
distributed capacitor is VCcT.

Current (I) is expressed as charge Q per unit time. Therefore the
current into the transmission line can be expressed as

I = Q/T =
VCcT / T = VCc

Impedance (Zo) is the ratio of voltage (V) to current (I). Therefore
the impedance can be expressed as

Zo = V / I =
V / VCc = 1/Cc

We can generalize this by using the velocity of the electrical wave
rather than the speed of light, which allows the formula to be applied
to transmission line with velocities slower than the speed of light.

Of course, only the wave front and wave end of a DC wave can be
measured to have a velocity.

73, Roger, W7WKB

In a 231 line posting that contains only original 57 lines:
On Thu, 13 Dec 2007 17:26:17 -0800, Roger wrote:

Hi Roger,

This last round has piqued my interest when we dipped into DC. Those
"formulas" would lead us to a DC wave velocity?

Hi Richard,

Here are two links to pages that cover the derivation of the formula Zo
= 1/cC and much more.

http://www.speedingedge.com/PDF-File..._Impedance.pdf
http://www.ece.uci.edu/docs/hspice/h...001_2-269.html

Here is the way I proposed to Kevin Schmidt nearly seven years ago after
seeing him use the formula on a web page:

Hi Roger,

However, none of what you respond with actually gives a DC wave
velocity. At a stretch, it is a transient with the potential of an
infinite number of waves (which could suffer dispersion from the
line's frequency characteristics making for an infinite number of
velocities). The infinite is a trivial observation in the scheme of
things when we return to DC.

Attaching a battery casts it into a role of AC generation (for however
long the transmission line takes to settle to an irresolvable
ringing). Discarding the term DC returns us to conventional
transmission line mechanics.

DC, in and of itself, has no wave velocity.

HTML enhanced text didn't help either. Some may shrug this off, but
browsers are the vectors of internet virii.

73's
Richard Clark, KB7QHC

"Roger" wrote in message
. ..
Richard Clark wrote:
On Thu, 13 Dec 2007 08:40:53 -0800, Roger wrote:


And just for completeness...
The fundamental equations also work when:
- the signal is not sinusoidal, e.g. pulse, step, square, ...
- rather than a load at one end, there is a source at each end
- the sources at each end produce different arbitrary functions
- the arbitrary functions at each end are DC sources
It is highly instructive to compute the forward and reverse
voltage and current (and then power) for a line with the same
DC voltage applied to each end.

...Keith

...Keith


Interesting! The important thing is to get answers that agree with
our experiments.

I have done some computations for DC voltage applied to transmission
lines. The real surprise for me came when I realized that transmission
line impedance could be expressed as a function of capacitance and the
wave velocity. Z0 = 1/cC where c is the velocity of the wave and C is
the capacitance of the transmission line per unit length.


Hi Roger,

This last round has piqued my interest when we dipped into DC. Those
"formulas" would lead us to a DC wave velocity?

73's
Richard Clark, KB7QHC

Hi Richard,

Here are two links to pages that cover the derivation of the formula Zo
= 1/cC and much more.

http://www.speedingedge.com/PDF-File..._Impedance.pdf
http://www.ece.uci.edu/docs/hspice/h...001_2-269.html

Here is the way I proposed to Kevin Schmidt nearly seven years ago after
seeing him use the formula on a web page:

*ASSUME*:
1) An electrical wave travels at the speed of light, c
2) A 'perfect' voltage source without impedance, V
3) A 'perfect' transmission line having no resistance but uniform
capacitance per unit length, C

*CONDITIONS AND SOLUTION*
The perfect voltage source has one terminal connected to the
transmission line prior to beginning the experiment. The experiment
begins by connecting the second terminal to the transmission line. The
voltage source drives an electrical wave down the transmission line at
the speed of light. Because of the limitation of speed, the wave
travels in the shape of a square wave containing all frequencies
required to create a square wave.

The square wave travels down the transmission line at the speed of light
(c). After time (T), the wave has traveled distance cT down the
transmission line, and has charged the distributed capacity CcT of the
line to voltage V over that distance. The total charge Q on the
distributed capacitor is VCcT.

Current (I) is expressed as charge Q per unit time. Therefore the
current into the transmission line can be expressed as

I = Q/T =
VCcT / T = VCc

Impedance (Zo) is the ratio of voltage (V) to current (I). Therefore
the impedance can be expressed as

Zo = V / I =
V / VCc = 1/Cc

We can generalize this by using the velocity of the electrical wave
rather than the speed of light, which allows the formula to be applied
to transmission line with velocities slower than the speed of light.

Of course, only the wave front and wave end of a DC wave can be
measured to have a velocity.

73, Roger, W7WKB

the OBVIOUS error is that the step when the second terminal is connected
DOES NOT travel down the line at c, it travels at some smaller percentage of
c given by the velocity factor of the line.

The second OBVIOUS error is the terminology 'DC wave'. you are measuring
the propagation velocity of a step function. this is a well defined fields
and waves 101 homework problem, not to be confused with the much more common
'sinusoidal stead state' solution that most other arguments on this group
assume but don't understand.

"Dave" wrote in message
news:q7u8j.6941$xd.2942@trndny03...

"Roger" wrote in message
. ..
Richard Clark wrote:
On Thu, 13 Dec 2007 08:40:53 -0800, Roger wrote:


And just for completeness...
The fundamental equations also work when:
- the signal is not sinusoidal, e.g. pulse, step, square, ...
- rather than a load at one end, there is a source at each end
- the sources at each end produce different arbitrary functions
- the arbitrary functions at each end are DC sources
It is highly instructive to compute the forward and reverse
voltage and current (and then power) for a line with the same
DC voltage applied to each end.

...Keith

...Keith


Interesting! The important thing is to get answers that agree with
our experiments.

I have done some computations for DC voltage applied to transmission
lines. The real surprise for me came when I realized that
transmission
line impedance could be expressed as a function of capacitance and the
wave velocity. Z0 = 1/cC where c is the velocity of the wave and C
is
the capacitance of the transmission line per unit length.


Hi Roger,

This last round has piqued my interest when we dipped into DC. Those
"formulas" would lead us to a DC wave velocity?

73's
Richard Clark, KB7QHC

Hi Richard,

Here are two links to pages that cover the derivation of the formula Zo
= 1/cC and much more.

http://www.speedingedge.com/PDF-File..._Impedance.pdf
http://www.ece.uci.edu/docs/hspice/h...001_2-269.html

Here is the way I proposed to Kevin Schmidt nearly seven years ago after
seeing him use the formula on a web page:

*ASSUME*:
1) An electrical wave travels at the speed of light, c
2) A 'perfect' voltage source without impedance, V
3) A 'perfect' transmission line having no resistance but uniform
capacitance per unit length, C

*CONDITIONS AND SOLUTION*
The perfect voltage source has one terminal connected to the
transmission line prior to beginning the experiment. The experiment
begins by connecting the second terminal to the transmission line. The
voltage source drives an electrical wave down the transmission line at
the speed of light. Because of the limitation of speed, the wave
travels in the shape of a square wave containing all frequencies
required to create a square wave.

The square wave travels down the transmission line at the speed of light
(c). After time (T), the wave has traveled distance cT down the
transmission line, and has charged the distributed capacity CcT of the
line to voltage V over that distance. The total charge Q on the
distributed capacitor is VCcT.

Current (I) is expressed as charge Q per unit time. Therefore the
current into the transmission line can be expressed as

I = Q/T =
VCcT / T = VCc

Impedance (Zo) is the ratio of voltage (V) to current (I). Therefore
the impedance can be expressed as

Zo = V / I =
V / VCc = 1/Cc

We can generalize this by using the velocity of the electrical wave
rather than the speed of light, which allows the formula to be applied
to transmission line with velocities slower than the speed of light.

Of course, only the wave front and wave end of a DC wave can be
measured to have a velocity.

73, Roger, W7WKB

the OBVIOUS error is that the step when the second terminal is connected
DOES NOT travel down the line at c, it travels at some smaller percentage
of c given by the velocity factor of the line.

That IS what I said. Think of the velocity as a moving wall, with the
capacitor charged behind the wall, uncharged in front of the moving wall.

The second OBVIOUS error is the terminology 'DC wave'. you are measuring
the propagation velocity of a step function. this is a well defined
fields and waves 101 homework problem, not to be confused with the much
more common 'sinusoidal stead state' solution that most other arguments on
this group assume but don't understand.



Be real. This experiment can be performed, and the DC switched as
frequently as desired. How square the wave front will be depends upon real
world factors.

Go to a transmission line characteristics table and use the formula to
compare Zo, capacity per length, and line velocity. It will amaze you.

73, Roger, W7WKB

AI4QJ wrote:
"Richard Clark" wrote in message
...
In a 231 line posting that contains only original 57 lines:
On Thu, 13 Dec 2007 17:26:17 -0800, Roger wrote:

Hi Roger,

This last round has piqued my interest when we dipped into DC. Those
"formulas" would lead us to a DC wave velocity?
Hi Richard,

Here are two links to pages that cover the derivation of the formula Zo
= 1/cC and much more.

http://www.speedingedge.com/PDF-File..._Impedance.pdf
http://www.ece.uci.edu/docs/hspice/h...001_2-269.html

Here is the way I proposed to Kevin Schmidt nearly seven years ago after
seeing him use the formula on a web page:
Hi Roger,

However, none of what you respond with actually gives a DC wave
velocity. At a stretch, it is a transient with the potential of an
infinite number of waves (which could suffer dispersion from the
line's frequency characteristics making for an infinite number of
velocities). The infinite is a trivial observation in the scheme of
things when we return to DC.

Attaching a battery casts it into a role of AC generation (for however
long the transmission line takes to settle to an irresolvable
ringing). Discarding the term DC returns us to conventional
transmission line mechanics.

DC, in and of itself, has no wave velocity.

For the model provided, R= 0, therefore we have a transmission line
consisting of superconductors. The speed at which steady state DC current is
injected into the model will equal the maximum speed of DC current in the
model. Although the electrons themselves will move very slowly, for each
coulomb injected in, one coulomb will be injected out at the same velocity
they were injected in (not to be confused with 'current' which is the number
of coulombs per second). If it were possible for the source to provide DC
current at c, then the DC current moves at c. The capacitance C can be any
value and Zo has no meaning. The only model that works here is the one with
a cardboard tube filled with ping pong balls, in this case with 0 distance
between them.

Ah, but of so little importance because the model is not reality.


While R (ohmic resistance) is specified as zero, impedance is what we
are looking for. Impedance is the ratio of voltage to current.

So far as we know, the maximum velocity permitted in the universe is the
speed of light, which is the speed of electromagnetic disturbance. Here
we disturb the transmission line electromagnetically. Think of the
velocity of the wave front as a moving wall. Everything behind the
moving wall is charged to the applied voltage, everything in front is
uncharged. The ratio of voltage to current turns out to be a pure
resistance, dependent only on the capacity per length and wave velocity.

Before becoming too critical or skeptical, run the equation with the
characteristics from a few transmission lines. You will find that the
numbers are very close, but not exact. The published characteristics
are not carried out to many decimal places, and who knows to what
accuracy they were determined.

73, Roger, W7WKB

AI4QJ wrote:
If it were possible for the source to provide DC
current at c, then the DC current moves at c.

The step function from zero to DC contains a lot
of frequencies. I suspect photons are involved
at the leading edge of the DC pulse.
--
73, Cecil http://www.w5dxp.com

On Dec 14, 9:40 am, Cecil Moore wrote:
AI4QJ wrote:
If it were possible for the source to provide DC
current at c, then the DC current moves at c.

The step function from zero to DC contains a lot
of frequencies. I suspect photons are involved
at the leading edge of the DC pulse.

"Suspect" -- Perhaps like Inspector Clouseau?

Humour aside, for transmission lines you should
stick to charge, and distributed capacitance and
inductance. This model is aptly capable and has
no difficulties as the frequency drops so low that
it becomes indistinguishable from DC.

Why bother with photons? Only at the leading
edge, you say. What explains the rest?

Where is the energy stored? In the capacitance
and inductance. Why not use the tools that work?

Why try to force fit photons?

....Keith

Keith Dysart wrote:
Why bother with photons?

Because it is impossible for electrons to move
fast enough to explain the measured results.
There is indeed a "DC" *wavefront* moving at
the speed of light adjusted for VF. Electrons
cannot move that fast. What is happening is
that fast photons are skipping from slow electron
to slow electron.
--
73, Cecil http://www.w5dxp.com