Standing-Wave Current vs Traveling-Wave Current
 

Art wrote:
"So I added a time varient to a static field which provided a direct
connection to Maxwell`s laws."

It is said that Maxwell began with Faraday`s law of electromagnetic
induction, which seems reasonable, as the voltage in one loop of wire
inductively coupled to another is proportional to the derivative dphi/dt
of current produced magnetic flux from the primary loop.

Induced voltage does not arise at a single point on the loop but is the
sum of all the infinitesimal bits of voltage distributed around the
loop. Maxwell developed his first field equation from this start.

Faraday`s law already deals with a time varying current to obtain
induction.

Different derivations are possible but Maxwell`s equations can express
all known electromagnetic conditions at once.

Oliver Heaviside made it his job to study Maxwell`s "Electromagnetic
Theory", learn its mathematics, and reduce it to vector analysis.
Heaviside published his results beginning a series of papers in 1887 in
a British publication, "The Electrician", in a very understandable way.
This has been called the start of modern communication engineering.

Art also wrote:
"The result for equilibrium must be a tank circuit where all energy is
interchanged between the energy storage tanks (inductance and
capacitance)"

True that an EM wave exchanges its energy between an electric field and
a magnetic field. Not true that a tank circuit type antenna is needed to
launch or capture EM waves. An isolated wire, resonant or non-resonant
is all that`s needed if it carries alternating current.

Nor are explosives needed to generate radio waves. Less disruptive
methods are quite acceptable.

Art also wrote:
"The bottom line is that for best efficiency for the unit volune
supplied to an array must be in equilibrium."

The above doesn`t square with what I`ve seen.
Efficiency is output over input. Antenna efficiency is usually Radiation
Resistance / Radiation Resistance + Loss Resistance. For the most
antenna for the money, Terman suggests the Yagi or the Corner Reflector.

Enough debunking for tonight.

Best regards, Richard Harrison, KB5WZI

Standing-Wave Current vs Traveling-Wave Current
 

"Richard Harrison" wrote in message
...
Art wrote:
"So I added a time varient to a static field which provided a direct
connection to Maxwell`s laws."

It is said that Maxwell began with Faraday`s law of electromagnetic

can we do this... Maxwell did not write 'laws'. all of my references refer
to "Maxwell's equations". Maxwell collected laws from Faraday, Gauss, Ohm,
Ampere, and probably a few others and as far as I know added just the
displacement current concept to turn them into a concise description of time
varying electromagnetic phenomenon that has never been disproved. This set
of equations is indeed what all antenna, feedline, waveguide, motors, power
transmission, and related electromagnetic phenomenon are modeled with
successfully in all environments that we have experienced. (neglecting
things like black holes anyway) This includes diamagnetic, paramagnetic,
ferromagnetic, insulators, conductors, dielectrics, and even the new man
made negative refractive index materials. it is not necessary to add a new
'time varient' part to any of Maxwell's equations, they already contain all
the time variant parts that are needed to explain any electromagnetic fields
you can come up with.

Standing-Wave Current vs Traveling-Wave Current
 

"Keith Dysart" wrote in message
...
Thanks for offering the two capacitor/one capacitor view of the middle
of the line. It took a bit of time to decide whether the commingling
of the charge in the single capacitor at the middle of the line would
solve my dilemma.

So I considered this one capacitor in the exact center of a perfect
transmission line. It is the perfect capacitor, absolutely
symmetrical. So as the exactly equal currents flow into it on
the exactly symmetrical leads, the charge is perfectly balanced
so that the charge coming from each side exactly occupies its
side of the conductor. As the two flows of charge flow over
the perfectly symmetrical plates, they meet in the exact
center, and flow no more. I conclude that a surface can
be found exactly in the center of this capacitor across
which no charge flows. Thus (un)happily returning me exactly
to where I was before; there is a line across which no
charge, and hence no energy, flows.

absolutely correct. no charge flows through the insulator of a capacitor.
you don't have to try hard to set up this condition, it is true of any
capacitor that has no resistive losses. the concept you are struggling with
is called the 'displacement current'. This is the unique concept that
Maxwell added to the laws of Gauss, Faraday, et al when he compiled his set
of electromagnetic equations. once you understand that then all will become
clear.

2. I don't understand the mechanism which causes waves to bounce.

I take this to imply that you are not happy with the simple "like
charge
repels"?


waves do not bounce. waves superimpose on each other, but the original
waves that meet in a homogenious, linear medium DO NOT INTERACT in any way.
they each proceed happily on their way without any change. it is only due
to the limited capability of your measurement instruments to measure the
superimposed field that results that causes the confusion. if you could
easily separate out the forward and reflected waves it would become
intuitively obvious... however the poor instruments (swr meters and other
'power' meters) that have caused the confusion and resulted in the misnomer
'standing wave'. It is of course not a 'wave', nor does it 'stand'... it is
just the result of superposition of 2 waves traveling in opposite
directions.



So the (poorly developped) "charge bouncing" explanation
seems like a way out, but I certainly would appreciate
other explanations for consideration.

you don't need a 'way out' you need to go back to basics and learn the
proper explanation.

A large part of the argument seems to revolve around a single point in a
perfect transmission line, where the current is exactly zero. This is an
infinitesimal point on a perfect line, so some anomalous things might be
expected to happen there.

but the current, or voltage, is only zero because of the superimposed
oppositely traveling waves. it is only your poor measurement instrument
that only responds to the superimposed waves that causes the confusion.


there. that ought to cause enough controversy to push this thread over the
1000 mark pretty quick! the wx is bad and i needed something to do today
anyway since i can't go out and work on my maxwell compliant antennas.

Standing-Wave Current vs Traveling-Wave Current
 

On 11 Jan, 05:00, "Dave" wrote:
"Richard Harrison" wrote in message

...

Art wrote:
"So I added a time varient to a static field which provided a direct
connection to Maxwell`s laws."

It is said that Maxwell began with Faraday`s law *of electromagnetic

can we do this... Maxwell did not write 'laws'. *all of my references refer
to "Maxwell's equations". *Maxwell collected laws from Faraday, Gauss, Ohm,
Ampere, and probably a few others and as far as I know added just the
displacement current concept to turn them into a concise description of time
varying electromagnetic phenomenon that has never been disproved. *This set
of equations is indeed what all antenna, feedline, waveguide, motors, power
transmission, and related electromagnetic phenomenon are modeled with
successfully in all environments that we have experienced. *(neglecting
things like black holes anyway) *This includes diamagnetic, paramagnetic,
ferromagnetic, insulators, conductors, dielectrics, and even the new man
made negative refractive index materials. *it is not necessary to add a new
'time varient' part to any of Maxwell's equations, they already contain all
the time variant parts that are needed to explain any electromagnetic fields
you can come up with.

I added a time varient to GAUSSIAN law to make it the same as
Maxwellian laws, not the other way around as you are inferring.
Before Maxwell came a long there were multiple laws stated many
different ways.
Maxwell saw that many of these laws tho written differently condensed
to the same thing.
So he condensed these multiple laws to a few but an no time did debunk
the laws that already were in situ. All of these multiple laws as with
Maxwell were built on the shoulders of Newton
and those that contributed before him, all of which are derived around
the pivot of gravity
and equilibrium

Standing-Wave Current vs Traveling-Wave Current
 

On Thu, 10 Jan 2008 18:52:28 -0800 (PST)
Keith Dysart wrote:
snip............

I would really appreciate seeing some other possible explanations.

One other one which I have seen and am not confortable with is the
explanation that energy in the waves pass through the point in
each direction and sum to zero. But this is indistinguishable from
superposing power which most agree is inappropriate. As well, this
explanation means that P(t) is not equal to V(t) times I(t),
something that I am quite reluctant to agree with.

The other explanation seen is that the voltage waves or the
current waves travel down the line superpose, yielding a total
voltage and current function at each point on the line which
can be used to compute the power. With this explanation, P(t)
is definitely equal to V(t) time I(t), which I do appreciate.
The weakness of this explanation is that it seems to deny
that the wave moves energy. And yet before the pulses collide
it is easy to observe the energy moving in the line, and if
a pulse was not coming in the other direction, there would
be no dispute that the energy travelled to the end of the
line and was absorbed in the load. Yet when the pulses
collide, no energy crosses the middle of the line. Yet
energy can be observed travelling in the line before
and after the pulses collide.

So...

I can give up on pulses (or waves) moving energy. I am not
happy doing that.
I can give up on P(t) = V(t) * I(t). I am not happy doing
that either.

So the (poorly developped) "charge bouncing" explanation
seems like a way out, but I certainly would appreciate
other explanations for consideration.

Hi Keith,

I think you are putting too much emphasis on the measurement of energy in the pulse, or maybe not enought, depending upon how you are making and using the measurements.

I hope you agree that the transmission line you are pulsing has no standing waves. If not, then the pulse must be a treveling wave. If it is a traveling wave, the impedance of the wave will be the impedance of the transmission line, except where the two pulses cross paths. At the crossing point, the impedance will jump to infinity, Z = v/i = v/zero = infinity.

How about the energy/power levels? Except at the center, as the pulse passes, current and voltage can be measured, so power is being delivered from one section of the line to a second section more distant from the source. The energy delivered by the power equation can be measured by U = (v^2)/r = (i^2)*r where r is the Zo if the transmission line.

Where the two pulses cross, the energy level must be the sum of the two energy levels, which would be 2U = 2*(v^2)/r (assuming two pulses of equal voltage). As you have pointed out, power would not flow at the point of wave first meeting, spreading from the meeting point back towards the two sources a distance equal to the square pulse width. Energy is present however, measured as U = (v^2)/r.

The transmission line can be considered as a lineal capacitor, so U = (CV^2)/2, where C is the capacitance per unit length. If the waves "bounce", and occupy physical space for a period of time, they should comingle, stop, and fill the capacitor during that time. If so,

2U = C*V'^2
= 2*(V^2)/r where v is the voltage of a single pulse and V' = 1,414*v = voltage observed at point of crossing.

At a disconinuity, we find a doubling of the voltages, not an increase of 1.414. This leads me to believe that the crossing pulses never stop in a physical sense. They do stop in the sense that power is not delived through the center point.

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

On 11 Jan, 08:15, Roger Sparks wrote:
On Thu, 10 Jan 2008 18:52:28 -0800 (PST)Keith Dysart wrote:

snip............







I would really appreciate seeing some other possible explanations.

One other one which I have seen and am not confortable with is the
explanation that energy in the waves pass through the point in
each direction and sum to zero. But this is indistinguishable from
superposing power which most agree is inappropriate. As well, this
explanation means that P(t) is not equal to V(t) times I(t),
something that I am quite reluctant to agree with.

The other explanation seen is that the voltage waves or the
current waves travel down the line superpose, yielding a total
voltage and current function at each point on the line which
can be used to compute the power. With this explanation, P(t)
is definitely equal to V(t) time I(t), which I do appreciate.
The weakness of this explanation is that it seems to deny
that the wave moves energy. And yet before the pulses collide
it is easy to observe the energy moving in the line, and if
a pulse was not coming in the other direction, there would
be no dispute that the energy travelled to the end of the
line and was absorbed in the load. Yet when the pulses
collide, no energy crosses the middle of the line. Yet
energy can be observed travelling in the line before
and after the pulses collide.

So...

I can give up on pulses (or waves) moving energy. I am not
happy doing that.
I can give up on P(t) = V(t) * I(t). I am not happy doing
that either.

So the (poorly developped) "charge bouncing" explanation
seems like a way out, but I certainly would appreciate
other explanations for consideration.

Hi Keith,

I think you are putting too much emphasis on the measurement of energy in the pulse, or maybe not enought, depending upon how you are making and using the measurements. *

I hope you agree that the transmission line you are pulsing has no standing waves. *If not, then the pulse must be a treveling wave. *If it is a traveling wave, the impedance of the wave will be the impedance of the transmission line, except where the two pulses cross paths. *At the crossing point, the impedance will jump to infinity, Z = v/i = v/zero = infinity.

How about the energy/power levels? *Except at the center, as the pulse passes, current and voltage can be measured, so power is being delivered from one section of the line to a second section more distant from the source. *The energy delivered by the power equation can be measured by U = (v^2)/r = (i^2)*r where r is the Zo if the transmission line.

Where the two pulses cross, the energy level must be the sum of the two energy levels, which would be 2U = 2*(v^2)/r (assuming two pulses of equal voltage). *As you have pointed out, power would not flow at the point of wave first meeting, spreading from the meeting point back towards the two sources a distance equal to the square pulse width. Energy is present however, measured as U = (v^2)/r.

The transmission line can be considered as a lineal capacitor, so U = (CV^2)/2, where C is the capacitance per unit length. *If the waves "bounce", and occupy physical space for a period of time, they should comingle, stop, and fill the capacitor during that time. If so,

2U = C*V'^2
* *= 2*(V^2)/r where v is the voltage of a single pulse and V' = 1,414*v = voltage observed at point of crossing. *

At a disconinuity, we find a doubling of the voltages, not an increase of 1.414. *This leads me to believe that the crossing pulses never stop in a physical sense. *They do stop in the sense that power is not delived through the center point.

73, Roger, W7WKB- Hide quoted text -

- Show quoted text -

Roger you are conversing with old men most of which oppose change.
Having been taught some fifty years ago they want to protect those
teachings the originators of which are all dead.
Thus regardless of what has happened and discovered or intruded upon
by inquiring minds they wish to stay in the past.In the past, statics
was viewed as a subset of electro-dynamics and the old timers want it
to stay that way. By starting with Gauss and adding a time varient I
added a visual to Maxwell that was missing from his equations because
Gauss was not the leading
contributor to the time varient theme. It is this that the old timers
are resisting because it brings away the thought that statics is just
a subset. Thus you see that there is inherrent resistance to the
mathematics that supplies a visual to Mawellian equations. It took
Richard several months to accept the legitimacy of the mathematics
involved in the steps I took. As yet nobody else has admitted to the
legitamacy of the mathematics. Thus protecting the notion of static
being just a subset. This manoever has placed ham radio into a
quagmire from which they cannot extricate themselves and thus advance
to analyse the intricacies of propagation.
Who would have believed that ham radio would take to task of a Doctor
working at MIT on space reseach for NASA for faulty use of mathematics
to prove the relavence of statics?
Can science really be held up while old men are still living?
Art Unwin KB9MZ...XG

Standing-Wave Current vs Traveling-Wave Current
 

roger chimed in with:

I hope you agree that the transmission line you are pulsing has no standing
waves. If not, then the pulse must be a treveling wave. If it is a
traveling wave, the impedance of the wave will be the impedance of the
transmission line, except where the two pulses cross paths. At the
crossing point, the impedance will jump to infinity, Z = v/i = v/zero =
infinity.


argh, another misconception!.. the impedance of a transmission line doesn't
change! and the impedance seen by a single traveling wave never changes
either. the measured ratio of voltage to current at a point on the line can
change as a result of the superposition of traveling waves. but again, that
is just a result of cheap instrumentation that can't resolve the component
waves.

Standing-Wave Current vs Traveling-Wave Current
 

On Fri, 11 Jan 2008 08:15:09 -0800, Roger Sparks
wrote:

They do stop in the sense that power is not delived through the center point.

Hi Roger,

I'm not sure if this is original to you, or one of those conceits that
is being passed around; but how, short of tagging energy (power?), can
you tell its origin? In other words, delivery by return, or
completion of the path needs to be separable by some information about
the origin that to this point has been missing from the discussion.

If energy can be tagged with a return address, what is that tag?

73's
Richard Clark, KB7QHC

Standing-Wave Current vs Traveling-Wave Current
 

On Fri, 11 Jan 2008 17:19:28 GMT
"Dave" wrote:

roger chimed in with:

I hope you agree that the transmission line you are pulsing has no standing
waves. If not, then the pulse must be a treveling wave. If it is a
traveling wave, the impedance of the wave will be the impedance of the
transmission line, except where the two pulses cross paths. At the
crossing point, the impedance will jump to infinity, Z = v/i = v/zero =
infinity.


argh, another misconception!.. the impedance of a transmission line doesn't
change! and the impedance seen by a single traveling wave never changes
either. the measured ratio of voltage to current at a point on the line can
change as a result of the superposition of traveling waves. but again, that
is just a result of cheap instrumentation that can't resolve the component
waves.


Good point. It would have been much better to say something like :"At the
crossing point, the MEASURED impedance will jump to infinity (Z = v/i = v/zero =
infinity) but we have no indication that the impedance of either the wire or the individual pulse changes in any way.".

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

On Fri, 11 Jan 2008 10:44:19 -0800
Richard Clark wrote:

On Fri, 11 Jan 2008 08:15:09 -0800, Roger Sparks
wrote:

They do stop in the sense that power is not delived through the center point.

Hi Roger,

I'm not sure if this is original to you, or one of those conceits that
is being passed around; but how, short of tagging energy (power?), can
you tell its origin? In other words, delivery by return, or
completion of the path needs to be separable by some information about
the origin that to this point has been missing from the discussion.

If energy can be tagged with a return address, what is that tag?

73's
Richard Clark, KB7QHC

Current is the tag, and the address is "past" and "future". Measurement is always "present", but we like to make predictions about what will happen in the future, and trace what has happened in the past. With the electromagnetic equations, we are quite good at making those extrapolations from present measurements.

At the crossing point, we can not measure current, so the power equation of P = V * I fails. There is no power at this point that we can measure. We can measure instantaneous voltage but don't know which equation to use to determine energy, U = (CV^2)/2 or SumU = Sum((V^2)/r). Only if we make many measurements over time can we learn the shape of the voltage over time, and thereby deduce which equation to use.

73, Roger, W7WKB






At the crossing poiont

Standing-Wave Current vs Traveling-Wave Current
 

On Fri, 11 Jan 2008 11:45:10 -0800, Roger Sparks
wrote:

Current is the tag
....
At the crossing point, we can not measure current,

Hi Roger,

Hence the information as to the source of energy that arrives at its
final destination is completely lost.

Barring this tracking, then, there is no way to prove effects due to
the "crossing point."

So, why does this discussion proceed without hope of any meaningful
proof? Certainly, given more than 400, 500, 1000 postings there must
be some other tagging mechanism that you are unaware of, or has been
withheld from comment. Did we miss that announcement?

73's
Richard Clark, KB7QHC

Standing-Wave Current vs Traveling-Wave Current
 

On Fri, 11 Jan 2008 08:44:01 -0800 (PST)
art wrote:


Roger you are conversing with old men most of which oppose change.
Having been taught some fifty years ago they want to protect those
teachings the originators of which are all dead.
Thus regardless of what has happened and discovered or intruded upon
by inquiring minds they wish to stay in the past.In the past, statics
was viewed as a subset of electro-dynamics and the old timers want it
to stay that way. By starting with Gauss and adding a time varient I
added a visual to Maxwell that was missing from his equations because
Gauss was not the leading
contributor to the time varient theme. It is this that the old timers
are resisting because it brings away the thought that statics is just
a subset. Thus you see that there is inherrent resistance to the
mathematics that supplies a visual to Mawellian equations. It took
Richard several months to accept the legitimacy of the mathematics
involved in the steps I took. As yet nobody else has admitted to the
legitamacy of the mathematics. Thus protecting the notion of static
being just a subset. This manoever has placed ham radio into a
quagmire from which they cannot extricate themselves and thus advance
to analyse the intricacies of propagation.
Who would have believed that ham radio would take to task of a Doctor
working at MIT on space reseach for NASA for faulty use of mathematics
to prove the relavence of statics?
Can science really be held up while old men are still living?
Art Unwin KB9MZ...XG

Hi Art,

Like you, I would like to find a better way of doing things, especially some easier math for the electromagnetic equations. Until we find that better way, we must use what we have.

What we have is very, very good, because it has brought us TV, computers, the Internet, space travel, and much more. What ever we find that could be "better", must also contain the knowledge we have today, but perhaps can explain todays knowledge easier or more completely.

I try to learn from the old men because the sum of their knowledge is far greater than my own can ever be. Having said that, I also recognize that the knowledge of each man is limited by his own talent, experience, and human frailty, so a judgement must be made as to wisdom and accuracy.

I really admire the willingness of all participants to contribute to discussions. The hard part is knowing how to fit each new tibit of knowledge into my own personal knowledge base.

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

On Fri, 11 Jan 2008 11:59:24 -0800
Richard Clark wrote:


So, why does this discussion proceed without hope of any meaningful
proof? Certainly, given more than 400, 500, 1000 postings there must
be some other tagging mechanism that you are unaware of, or has been
withheld from comment. Did we miss that announcement?

73's
Richard Clark, KB7QHC

The writtings of many of us still have "hope". Many postings reflect a curiosity that has not been satisfied. Is there a tagging mechanism that I did not mention? If so, please articulate it.

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

On Fri, 11 Jan 2008 12:23:39 -0800, Roger Sparks
wrote:

The writtings of many of us still have "hope".

Hi Roger,

Ah, yes. Hope without proof is called faith. There are testimonials,
and there are witnesses to miracles - however, those, too, seem to be
far and few between.

Faith in "what" seems to be an even greater mystery, and a strange
basis for belief.

Another conceit I've seen ripple through the threads - I don't think
you offered it either, but authorship is vague - is that this "what"
may be waves bouncing.

Hope in waves bouncing (for which only faith supports this where proof
is absent) may solve the mystery of "what," but like Russian Dolls
nested one inside another, certainly this faith is ever spiraling out
toward some even greater mystery in beliefs.

To put it in religious terms: "So What?"

Is this the road to redemption?

Is it like the discovery of the mirror? Did the discoverer (inventor
if you will seeing as the mirror might be currently in the process of
patenting) attain a new insight? a new perspective?

Even there, our using (measuring with) a mirror gives us a means to
tag information through parallax. Is this thread missing something as
simple as parallax to tag currents as to their origin? It would seem
after 800 contributions that this contribution (parallax) would have
resolved the thread and brought closure to allow everyone to transcend
obstacles on that road to redemption.

Is anyone actually pursuing the goal of finding and expressing the
parallax problem (or revealing any tagging); or are they simply
content to have hope and wait for the end of days milling about the
edges of the purgatory of Babel?

73's
Richard Clark, KB7QHC

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 11, 8:12*am, "Dave" wrote:
"Keith Dysart" wrote in message

...

Thanks for offering the two capacitor/one capacitor view of the middle
of the line. It took a bit of time to decide whether the commingling
of the charge in the single capacitor at the middle of the line would
solve my dilemma.

So I considered this one capacitor in the exact center of a perfect
transmission line. It is the perfect capacitor, absolutely
symmetrical. So as the exactly equal currents flow into it on
the exactly symmetrical leads, the charge is perfectly balanced
so that the charge coming from each side exactly occupies its
side of the conductor. As the two flows of charge flow over
the perfectly symmetrical plates, they meet in the exact
center, and flow no more. I conclude that a surface can
be found *exactly in the center of this capacitor across
which no charge flows. Thus (un)happily returning me exactly
to where I was before; there is a line across which no
charge, and hence no energy, flows.

absolutely correct. *no charge flows through the insulator of a capacitor.

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 11, 11:15*am, Roger Sparks wrote:
On Thu, 10 Jan 2008 18:52:28 -0800 (PST)Keith Dysart wrote:

snip............



I would really appreciate seeing some other possible explanations.

One other one which I have seen and am not confortable with is the
explanation that energy in the waves pass through the point in
each direction and sum to zero. But this is indistinguishable from
superposing power which most agree is inappropriate. As well, this
explanation means that P(t) is not equal to V(t) times I(t),
something that I am quite reluctant to agree with.

The other explanation seen is that the voltage waves or the
current waves travel down the line superpose, yielding a total
voltage and current function at each point on the line which
can be used to compute the power. With this explanation, P(t)
is definitely equal to V(t) time I(t), which I do appreciate.
The weakness of this explanation is that it seems to deny
that the wave moves energy. And yet before the pulses collide
it is easy to observe the energy moving in the line, and if
a pulse was not coming in the other direction, there would
be no dispute that the energy travelled to the end of the
line and was absorbed in the load. Yet when the pulses
collide, no energy crosses the middle of the line. Yet
energy can be observed travelling in the line before
and after the pulses collide.

So...

I can give up on pulses (or waves) moving energy. I am not
happy doing that.
I can give up on P(t) = V(t) * I(t). I am not happy doing
that either.

So the (poorly developped) "charge bouncing" explanation
seems like a way out, but I certainly would appreciate
other explanations for consideration.

Hi Keith,

I think you are putting too much emphasis on the measurement of energy in the pulse, or maybe not enought, depending upon how you are making and using the measurements. *

I hope you agree that the transmission line you are pulsing has no standing waves. *

"Standing waves" seem to more be a consequence of sinusoidal
excitation than pulse.
So I expect no "standing waves".

I don't even like using the words "traveling wave" to describe the
pulse situation
since "traveling wave" also seems to bring a lot of baggage.

If not, then the pulse must be a treveling wave. *If it is a traveling wave, the impedance of the wave will be the impedance of the transmission line, except where the two pulses cross paths. *At the crossing point, the impedance will jump to infinity, Z = v/i = v/zero = infinity.

I do not recommend computing this ratio and calling it impedance. I
find it leads
to difficulties. Better just to think of the V and I independantly.

How about the energy/power levels? *Except at the center, as the pulse passes, current and voltage can be measured, so power is being delivered from one section of the line to a second section more distant from the source. *The energy delivered by the power equation can be measured by U = (v^2)/r = (i^2)*r where r is the Zo if the transmission line.

More accurately P(t) = V(t) * I(t). it happens that for the pulses in
question,
except when colliding, that V and I of the pulses observed on the line
are related
by Z0 so by substitution, the expressions above can be derived. But
sometimes it is
better to stick with the more basic expression P=VI.

Where the two pulses cross, the energy level must be the sum of the two energy levels, which would be 2U = 2*(v^2)/r (assuming two pulses of equal voltage). *As you have pointed out, power would not flow at the point of wave first meeting, spreading from the meeting point back towards the two sources a distance equal to the square pulse width. Energy is present however, measured as U = (v^2)/r.

The transmission line can be considered as a lineal capacitor, so U = (CV^2)/2, where C is the capacitance per unit length. *If the waves "bounce", and occupy physical space for a period of time, they should comingle, stop, and fill the capacitor during that time. If so,

2U = C*V'^2
* *= 2*(V^2)/r where v is the voltage of a single pulse and V' = 1,414*v = voltage observed at point of crossing. *

But the voltage at the crossing is 2V. A puzzle? When the pulse is
"travelling",
there is energy in the capacitance and in the inductance of the line.
When the
current stops, the energy in the inductance is added to the energy in
the
capacitance, thus the voltage reaches 2V.

At a disconinuity, we find a doubling of the voltages, not an increase of 1.414. *This leads me to believe that the crossing pulses never stop in a physical sense. *They do stop in the sense that power is not delived through the center point.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 11, 1:44*pm, Richard Clark wrote:
On Fri, 11 Jan 2008 08:15:09 -0800, Roger Sparks
wrote:

They do stop in the sense that power is not delived through the center point.

Hi Roger,

I'm not sure if this is original to you, or one of those conceits that
is being passed around; but how, short of tagging energy (power?), can
you tell its origin? *In other words, delivery by return, or
completion of the path needs to be separable by some information about
the origin that to this point has been missing from the discussion.

If energy can be tagged with a return address, what is that tag?

I suggest that since the current at the middle of the line
is always 0, the power is always 0 and therefore no energy
crosses the middle. While tracking mixed energy is difficult,
when no energy crosses a boundary it seems easy to keep the
energy on each side of the boundary separate.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

"Keith Dysart" wrote in message
...

absolutely correct. no charge flows through the insulator of a capacitor.

This was not the line I meant. My line was perpendicular to the
plates. Current does not cross this line because it arrives equally from
both
sides.

huh?? no charge passes from one plate of a capacitor to the other. but
current does in the form of displacement current. if you are trying to look
at the microscopic field and charge distributions during a transient in a
capacitor you are way beyond anything i ever studied, or would want to
study.

waves do not bounce.

What about charge?

charge is being moved by the fields of the wave, and it moves very little in
the wire as has been shown in the past. when equal and opposite waves
arrived at a point where the current cancels the charges superimpose to
result in a zero field so there is no motion, hence no current at that
point. as the waves pass each other the fields no longer cancel so the
electrons on the other side of the cancelation point start to move again in
the opposite directions. if you want to call that a bounce knock yourself
out, but it will lead to a rather limited view that is only helpful in
certain circumstances where exact cancelation occurs.

Or is it the "standing wave" which is real and it just happens that
it can also be described as the sum of two waves going in opposite
directions? But the wave going in one direction can be described
as the sum of a number of waves going in that direction? (Especially\
when there are multiple reflections!) How many waves in each
direction?
When do we stop?

I respectfully submit that it is only the resultant that is real.
The rest are merely alternate descriptions (though often useful).

then enjoy your fantasy. the forward and reverse wave components are easily
separated and observed anywhere on the line. its just that most amateur
grade equipment doesn't bother because it is sufficient for 99.44% of us to
only measure your 'real' standing wave to properly adjust our antennas and
radios.

Standing-Wave Current vs Traveling-Wave Current
 

On Fri, 11 Jan 2008 14:03:10 -0800 (PST), Keith Dysart
wrote:

On Jan 11, 1:44*pm, Richard Clark wrote:
On Fri, 11 Jan 2008 08:15:09 -0800, Roger Sparks
wrote:

They do stop in the sense that power is not delived through the center point.

Hi Roger,

I'm not sure if this is original to you, or one of those conceits that
is being passed around; but how, short of tagging energy (power?), can
you tell its origin? *In other words, delivery by return, or
completion of the path needs to be separable by some information about
the origin that to this point has been missing from the discussion.

If energy can be tagged with a return address, what is that tag?

I suggest that since the current at the middle of the line
is always 0, the power is always 0 and therefore no energy
crosses the middle. While tracking mixed energy is difficult,
when no energy crosses a boundary it seems easy to keep the
energy on each side of the boundary separate.

Hi Keith,

If you cannot identify the source, then you cannot proclaim a
separation.

The best term for this is "the enigma effect." I hate to turn this
loose on the world because I know many here will pick it up and run to
their perceived goal for a touch-down.

Before anyone tackles me for a touchback, I'm going to pass this off
and watch the quarterbacks wrestle with religion in the huddle.

73's
Richard Clark, KB7QHC

Standing-Wave Current vs Traveling-Wave Current
 

On 11 Jan, 14:03, Keith Dysart wrote:
On Jan 11, 1:44*pm, Richard Clark wrote:

On Fri, 11 Jan 2008 08:15:09 -0800, Roger Sparks
wrote:

They do stop in the sense that power is not delived through the center point.

Hi Roger,

I'm not sure if this is original to you, or one of those conceits that
is being passed around; but how, short of tagging energy (power?), can
you tell its origin? *In other words, delivery by return, or
completion of the path needs to be separable by some information about
the origin that to this point has been missing from the discussion.

If energy can be tagged with a return address, what is that tag?

I suggest that since the current at the middle of the line
is always 0, the power is always 0 and therefore no energy
crosses the middle. While tracking mixed energy is difficult,
when no energy crosses a boundary it seems easy to keep the
energy on each side of the boundary separate.

...Keith

Keith,
Richard has not entered the actual discussion nor has he shown any
indication
he understands what is going on. All he is doing is intimating he
knows all the answers but he is not going to reveal them. Why? He has
no choice if later he wants to state "I knew that all the time". In
the mean time he is taunting people with off subject topics and then
deigns not to provide technical content. He is very good at arousing
people to respond to him in a defensive manner as you can now
personally point out, usually by asking questions without any effort
to support the discussion.His degree is in english which is why he is
practicing on this newsgroup
Have a great week end
Art

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 11, 5:03 pm, Keith Dysart wrote:
I suggest that since the current at the middle of the line
is always 0, the power is always 0 and therefore no energy
crosses the middle. While tracking mixed energy is difficult,
when no energy crosses a boundary it seems easy to keep the
energy on each side of the boundary separate.

Of course, what you say is only true for net energy. If we have a
forward Poynting vector of 100 watts and a reflected Poynting vector
of 100 watts, no net energy is flowing at any point.
--
73, Cecil, w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Richard Clark wrote:

On Fri, 11 Jan 2008 14:03:10 -0800 (PST), Keith Dysart
wrote:


On Jan 11, 1:44 pm, Richard Clark wrote:

On Fri, 11 Jan 2008 08:15:09 -0800, Roger Sparks
wrote:


They do stop in the sense that power is not delived through the center point.

Hi Roger,

I'm not sure if this is original to you, or one of those conceits that
is being passed around; but how, short of tagging energy (power?), can
you tell its origin? In other words, delivery by return, or
completion of the path needs to be separable by some information about
the origin that to this point has been missing from the discussion.

If energy can be tagged with a return address, what is that tag?

I suggest that since the current at the middle of the line
is always 0, the power is always 0 and therefore no energy
crosses the middle. While tracking mixed energy is difficult,
when no energy crosses a boundary it seems easy to keep the
energy on each side of the boundary separate.


Hi Keith,

If you cannot identify the source, then you cannot proclaim a
separation.

The best term for this is "the enigma effect." I hate to turn this
loose on the world because I know many here will pick it up and run to
their perceived goal for a touch-down.

Before anyone tackles me for a touchback, I'm going to pass this off
and watch the quarterbacks wrestle with religion in the huddle.

73's
Richard Clark, KB7QHC

To single out a particular spot as the site of a lack of energy
movement along an equipotential system in which there is no tranfer of
energy anyway seems to me to be an arbitrary convenience.

Maxwell explained how electromagentic waves 'bounce' if they encounter
matter, and I haven't seen any evidence that it was an incomplete
description.

ac6xg

Standing-Wave Current vs Traveling-Wave Current
 

On Fri, 11 Jan 2008 13:59:47 -0800 (PST)
Keith Dysart wrote:

On Jan 11, 11:15*am, Roger Sparks wrote:
On Thu, 10 Jan 2008 18:52:28 -0800 (PST)Keith Dysart wrote:

snip............



I would really appreciate seeing some other possible explanations.

One other one which I have seen and am not confortable with is the
explanation that energy in the waves pass through the point in
each direction and sum to zero. But this is indistinguishable from
superposing power which most agree is inappropriate. As well, this
explanation means that P(t) is not equal to V(t) times I(t),
something that I am quite reluctant to agree with.

The other explanation seen is that the voltage waves or the
current waves travel down the line superpose, yielding a total
voltage and current function at each point on the line which
can be used to compute the power. With this explanation, P(t)
is definitely equal to V(t) time I(t), which I do appreciate.
The weakness of this explanation is that it seems to deny
that the wave moves energy. And yet before the pulses collide
it is easy to observe the energy moving in the line, and if
a pulse was not coming in the other direction, there would
be no dispute that the energy travelled to the end of the
line and was absorbed in the load. Yet when the pulses
collide, no energy crosses the middle of the line. Yet
energy can be observed travelling in the line before
and after the pulses collide.

So...

I can give up on pulses (or waves) moving energy. I am not
happy doing that.
I can give up on P(t) = V(t) * I(t). I am not happy doing
that either.

So the (poorly developped) "charge bouncing" explanation
seems like a way out, but I certainly would appreciate
other explanations for consideration.

Hi Keith,

I think you are putting too much emphasis on the measurement of energy in the pulse, or maybe not enought, depending upon how you are making and using the measurements. *

I hope you agree that the transmission line you are pulsing has no standing waves. *

"Standing waves" seem to more be a consequence of sinusoidal
excitation than pulse.
So I expect no "standing waves".

I don't even like using the words "traveling wave" to describe the
pulse situation
since "traveling wave" also seems to bring a lot of baggage.

If not, then the pulse must be a treveling wave. *If it is a traveling wave, the impedance of the wave will be the impedance of the transmission line, except where the two pulses cross paths. *At the crossing point, the impedance will jump to infinity, Z = v/i = v/zero = infinity.

I do not recommend computing this ratio and calling it impedance. I
find it leads
to difficulties. Better just to think of the V and I independantly.

How about the energy/power levels? *Except at the center, as the pulse passes, current and voltage can be measured, so power is being delivered from one section of the line to a second section more distant from the source. *The energy delivered by the power equation can be measured by U = (v^2)/r = (i^2)*r where r is the Zo if the transmission line.

More accurately P(t) = V(t) * I(t). it happens that for the pulses in
question,
except when colliding, that V and I of the pulses observed on the line
are related
by Z0 so by substitution, the expressions above can be derived. But
sometimes it is
better to stick with the more basic expression P=VI.

Where the two pulses cross, the energy level must be the sum of the two energy levels, which would be 2U = 2*(v^2)/r (assuming two pulses of equal voltage). *As you have pointed out, power would not flow at the point of wave first meeting, spreading from the meeting point back towards the two sources a distance equal to the square pulse width. Energy is present however, measured as U = (v^2)/r.

The transmission line can be considered as a lineal capacitor, so U = (CV^2)/2, where C is the capacitance per unit length. *If the waves "bounce", and occupy physical space for a period of time, they should comingle, stop, and fill the capacitor during that time. If so,

2U = C*V'^2
* *= 2*(V^2)/r where v is the voltage of a single pulse and V' = 1,414*v = voltage observed at point of crossing. *

But the voltage at the crossing is 2V. A puzzle? When the pulse is
"travelling",
there is energy in the capacitance and in the inductance of the line.
When the
current stops, the energy in the inductance is added to the energy in
the
capacitance, thus the voltage reaches 2V.

Which is correct? P = (v^2)/r = (i^2)r or P = (v^2)/r + (i^2)r

If P = (v^2)/r = (i^2)r is correct, what is the precondition that MUST be true?
Answer: The i and v measurements must be across a pure resistance, without reactance.

The traveling wave is considered to have a pure resistive impedance.

This going by measurements makes things hard! Too bad we can't see and feel the wave with our own eyes and hands!


At a disconinuity, we find a doubling of the voltages, not an increase of 1.414. *This leads me to believe that the crossing pulses never stop in a physical sense. *They do stop in the sense that power is not delived through the center point.

...Keith

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 11, 7:30*pm, Roger Sparks wrote:
On Fri, 11 Jan 2008 13:59:47 -0800 (PST)





Keith Dysart wrote:
On Jan 11, 11:15*am, Roger Sparks wrote:
On Thu, 10 Jan 2008 18:52:28 -0800 (PST)Keith Dysart wrote:

snip............

I would really appreciate seeing some other possible explanations.

One other one which I have seen and am not confortable with is the
explanation that energy in the waves pass through the point in
each direction and sum to zero. But this is indistinguishable from
superposing power which most agree is inappropriate. As well, this
explanation means that P(t) is not equal to V(t) times I(t),
something that I am quite reluctant to agree with.

The other explanation seen is that the voltage waves or the
current waves travel down the line superpose, yielding a total
voltage and current function at each point on the line which
can be used to compute the power. With this explanation, P(t)
is definitely equal to V(t) time I(t), which I do appreciate.
The weakness of this explanation is that it seems to deny
that the wave moves energy. And yet before the pulses collide
it is easy to observe the energy moving in the line, and if
a pulse was not coming in the other direction, there would
be no dispute that the energy travelled to the end of the
line and was absorbed in the load. Yet when the pulses
collide, no energy crosses the middle of the line. Yet
energy can be observed travelling in the line before
and after the pulses collide.

So...

I can give up on pulses (or waves) moving energy. I am not
happy doing that.
I can give up on P(t) = V(t) * I(t). I am not happy doing
that either.

So the (poorly developped) "charge bouncing" explanation
seems like a way out, but I certainly would appreciate
other explanations for consideration.

Hi Keith,

I think you are putting too much emphasis on the measurement of energy in the pulse, or maybe not enought, depending upon how you are making and using the measurements. *

I hope you agree that the transmission line you are pulsing has no standing waves. *

"Standing waves" seem to more be a consequence of sinusoidal
excitation than pulse.
So I expect no "standing waves".

I don't even like using the words "traveling wave" to describe the
pulse situation
since "traveling wave" also seems to bring a lot of baggage.

If not, then the pulse must be a treveling wave. *If it is a traveling wave, the impedance of the wave will be the impedance of the transmission line, except where the two pulses cross paths. *At the crossing point, the impedance will jump to infinity, Z = v/i = v/zero = infinity.

I do not recommend computing this ratio and calling it impedance. I
find it leads
to difficulties. Better just to think of the V and I independantly.

How about the energy/power levels? *Except at the center, as the pulse passes, current and voltage can be measured, so power is being delivered from one section of the line to a second section more distant from the source. *The energy delivered by the power equation can be measured by U = (v^2)/r = (i^2)*r where r is the Zo if the transmission line.

More accurately P(t) = V(t) * I(t). it happens that for the pulses in
question,
except when colliding, that V and I of the pulses observed on the line
are related
by Z0 so by substitution, the expressions above can be derived. But
sometimes it is
better to stick with the more basic expression P=VI.

Where the two pulses cross, the energy level must be the sum of the two energy levels, which would be 2U = 2*(v^2)/r (assuming two pulses of equal voltage). *As you have pointed out, power would not flow at the point of wave first meeting, spreading from the meeting point back towards the two sources a distance equal to the square pulse width. Energy is present however, measured as U = (v^2)/r.

The transmission line can be considered as a lineal capacitor, so U = (CV^2)/2, where C is the capacitance per unit length. *If the waves "bounce", and occupy physical space for a period of time, they should comingle, stop, and fill the capacitor during that time. If so,

2U = C*V'^2
* *= 2*(V^2)/r where v is the voltage of a single pulse and V' = 1,414*v = voltage observed at point of crossing. *

But the voltage at the crossing is 2V. A puzzle? When the pulse is
"travelling",
there is energy in the capacitance and in the inductance of the line.
When the
current stops, the energy in the inductance is added to the energy in
the
capacitance, thus the voltage reaches 2V.

Which is correct? *P = (v^2)/r = (i^2)r * or * P = (v^2)/r + (i^2)r

P(t) = V(t) * I(t)
by substitution from V = I * R
P(t) = V(t) * V(t) / R
and
P(t) = I(t) * I(t) * R

But recall that power is the flow of energy per unit time.

The energy present in any section of line is the energy
present in the electric field of the capacitance plus
the energy present in the magnetic field of the inductance.

So from your expressions above
U = (CV^2)/2 + (LI^2)/2

If P = (v^2)/r = (i^2)r is correct, what is the precondition that MUST be true?
Answer: The i and v measurements must be across a pure resistance, without reactance.

This is why it is often better just to use the original
equation:
P(t) = V(t) * I(t)
for it is always true and requires no preconditions.

The traveling wave is considered to have a pure resistive impedance.

This going by measurements makes things hard! *Too bad we can't see and feel the wave with our own eyes and hands!



At a disconinuity, we find a doubling of the voltages, not an increase of 1.414. *This leads me to believe that the crossing pulses never stop in a physical sense. *They do stop in the sense that power is not delived through the center point.

...Keith

73, Roger, W7WKB- Hide quoted text -

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 11, 5:25*pm, Richard Clark wrote:
On Fri, 11 Jan 2008 14:03:10 -0800 (PST), Keith Dysart





wrote:
On Jan 11, 1:44*pm, Richard Clark wrote:
On Fri, 11 Jan 2008 08:15:09 -0800, Roger Sparks
wrote:

They do stop in the sense that power is not delived through the center point.

Hi Roger,

I'm not sure if this is original to you, or one of those conceits that
is being passed around; but how, short of tagging energy (power?), can
you tell its origin? *In other words, delivery by return, or
completion of the path needs to be separable by some information about
the origin that to this point has been missing from the discussion.

If energy can be tagged with a return address, what is that tag?

I suggest that since the current at the middle of the line
is always 0, the power is always 0 and therefore no energy
crosses the middle. While tracking mixed energy is difficult,
when no energy crosses a boundary it seems easy to keep the
energy on each side of the boundary separate.

Hi Keith,

If you cannot identify the source, then you cannot proclaim a
separation.

That is hardly difficult. If no energy every crosses
the middle of the line, then the energy on the left half
of the line originated with the source on the left side
and ditto for the right.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

On Fri, 11 Jan 2008 18:51:07 -0800 (PST), Keith Dysart
wrote:

If you cannot identify the source, then you cannot proclaim a
separation.

That is hardly difficult. If no energy every crosses
the middle of the line,

Hi Keith,

If indeed.

The difficulty is you've already allowed that the energy is not
tagged, so its source is unknown at the load. So "if" fails through
this immutable correlation. The separation is not demonstrated.

73's
Richard Clark, KB7QHC

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 11, 6:56*pm, Cecil Moore wrote:
On Jan 11, 5:03 pm, Keith Dysart wrote:

I suggest that since the current at the middle of the line
is always 0, the power is always 0 and therefore no energy
crosses the middle. While tracking mixed energy is difficult,
when no energy crosses a boundary it seems easy to keep the
energy on each side of the boundary separate.

Of course, what you say is only true for net energy. If we have a
forward Poynting vector of 100 watts and a reflected Poynting vector
of 100 watts, no net energy is flowing at any point.

You don't need Poynting vectors to realize that when
the instantaneous power is always 0, no energy is flowing.
And when the instantaneous power is always 0, it is
unnecessary to integrate and average to compute the
net energy flow, because no energy is flowing at all.

And if by your response you really do mean that energy
can be flowing when the instantaneous power is always 0,
please be direct and say so.

But then you will have to come up with a new definition
of instantaneous power for it can not be that it is
the rate of energy transfer if energy is flowing when
the instantaneous power is zero.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

On Fri, 11 Jan 2008 18:49:20 -0800 (PST)
Keith Dysart wrote:

On Jan 11, 7:30*pm, Roger Sparks wrote:
On Fri, 11 Jan 2008 13:59:47 -0800 (PST)





Keith Dysart wrote:
On Jan 11, 11:15*am, Roger Sparks wrote:
On Thu, 10 Jan 2008 18:52:28 -0800 (PST)Keith Dysart wrote:

snip............

I would really appreciate seeing some other possible explanations.

One other one which I have seen and am not confortable with is the
explanation that energy in the waves pass through the point in
each direction and sum to zero. But this is indistinguishable from
superposing power which most agree is inappropriate. As well, this
explanation means that P(t) is not equal to V(t) times I(t),
something that I am quite reluctant to agree with.

The other explanation seen is that the voltage waves or the
current waves travel down the line superpose, yielding a total
voltage and current function at each point on the line which
can be used to compute the power. With this explanation, P(t)
is definitely equal to V(t) time I(t), which I do appreciate.
The weakness of this explanation is that it seems to deny
that the wave moves energy. And yet before the pulses collide
it is easy to observe the energy moving in the line, and if
a pulse was not coming in the other direction, there would
be no dispute that the energy travelled to the end of the
line and was absorbed in the load. Yet when the pulses
collide, no energy crosses the middle of the line. Yet
energy can be observed travelling in the line before
and after the pulses collide.

So...

I can give up on pulses (or waves) moving energy. I am not
happy doing that.
I can give up on P(t) = V(t) * I(t). I am not happy doing
that either.

So the (poorly developped) "charge bouncing" explanation
seems like a way out, but I certainly would appreciate
other explanations for consideration.

Hi Keith,

I think you are putting too much emphasis on the measurement of energy in the pulse, or maybe not enought, depending upon how you are making and using the measurements. *

I hope you agree that the transmission line you are pulsing has no standing waves. *

"Standing waves" seem to more be a consequence of sinusoidal
excitation than pulse.
So I expect no "standing waves".

I don't even like using the words "traveling wave" to describe the
pulse situation
since "traveling wave" also seems to bring a lot of baggage.

If not, then the pulse must be a treveling wave. *If it is a traveling wave, the impedance of the wave will be the impedance of the transmission line, except where the two pulses cross paths. *At the crossing point, the impedance will jump to infinity, Z = v/i = v/zero = infinity.

I do not recommend computing this ratio and calling it impedance. I
find it leads
to difficulties. Better just to think of the V and I independantly.

How about the energy/power levels? *Except at the center, as the pulse passes, current and voltage can be measured, so power is being delivered from one section of the line to a second section more distant from the source. *The energy delivered by the power equation can be measured by U = (v^2)/r = (i^2)*r where r is the Zo if the transmission line.

More accurately P(t) = V(t) * I(t). it happens that for the pulses in
question,
except when colliding, that V and I of the pulses observed on the line
are related
by Z0 so by substitution, the expressions above can be derived. But
sometimes it is
better to stick with the more basic expression P=VI.

Where the two pulses cross, the energy level must be the sum of the two energy levels, which would be 2U = 2*(v^2)/r (assuming two pulses of equal voltage). *As you have pointed out, power would not flow at the point of wave first meeting, spreading from the meeting point back towards the two sources a distance equal to the square pulse width. Energy is present however, measured as U = (v^2)/r.

The transmission line can be considered as a lineal capacitor, so U = (CV^2)/2, where C is the capacitance per unit length. *If the waves "bounce", and occupy physical space for a period of time, they should comingle, stop, and fill the capacitor during that time. If so,

2U = C*V'^2
* *= 2*(V^2)/r where v is the voltage of a single pulse and V' = 1,414*v = voltage observed at point of crossing. *

But the voltage at the crossing is 2V. A puzzle? When the pulse is
"travelling",
there is energy in the capacitance and in the inductance of the line.
When the
current stops, the energy in the inductance is added to the energy in
the
capacitance, thus the voltage reaches 2V.

Which is correct? *P = (v^2)/r = (i^2)r * or * P = (v^2)/r + (i^2)r

P(t) = V(t) * I(t)
by substitution from V = I * R
P(t) = V(t) * V(t) / R
and
P(t) = I(t) * I(t) * R

Correct.

But recall that power is the flow of energy per unit time.

The energy present in any section of line is the energy
present in the electric field of the capacitance plus
the energy present in the magnetic field of the inductance.

When the capacitance is in phase with the magnetic field, both are measured to peak at the same time. This "resistive" condition is found in a transmission line when there are no standing waves. Under resistive conditions, the electric field and inductive field merge into one electromagnetic field that carries only one energy. The identical charges carry both electric and magnetic fields, with the fields in a ratio defined as impedance. Each field is a separate way of looking at the same energy.

So from your expressions above
U = (CV^2)/2 + (LI^2)/2

You may have been reading about traveling waves. They offer an explaination for how magnetic and electric fields may get out of phase. When the two fields are out of phase, the U = (CV^2)/2 + (LI^2)/2 calculation will ALWAYS sum to more energy than can be found in the resistive wave. This occurs because more energy actually is resident on the line with reflections, due to waves traveling in two directions. The reflection effectively doubles the length of the transmission line available for energy storage.


If P = (v^2)/r = (i^2)r is correct, what is the precondition that MUST be true?
Answer: The i and v measurements must be across a pure resistance, without reactance.

This is why it is often better just to use the original
equation:
P(t) = V(t) * I(t)
for it is always true and requires no preconditions.

I learned long ago that it DOES have a precondition. The power measured will only be resistive if measured across a pure resistance. If the measurement is across resistance plus reactance, then reactive power is also measured so P(t) = V(t) * I(t) becomes the sum of an unknown mix of two or more waves.


The traveling wave is considered to have a pure resistive impedance.

This going by measurements makes things hard! *Too bad we can't see and feel the wave with our own eyes and hands!



At a disconinuity, we find a doubling of the voltages, not an increase of 1.414. *This leads me to believe that the crossing pulses never stop in a physical sense. *They do stop in the sense that power is not delived through the center point.

...Keith

73, Roger, W7WKB- Hide quoted text -

...Keith

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 11, 10:06*pm, Richard Clark wrote:
On Fri, 11 Jan 2008 18:51:07 -0800 (PST), Keith Dysart

wrote:
If you cannot identify the source, then you cannot proclaim a
separation.

That is hardly difficult. If no energy every crosses
the middle of the line,

Hi Keith,

If indeed. *

The difficulty is *you've already allowed that the energy is not
tagged, so its source is unknown at the load. *So "if" fails through
this immutable correlation. *The separation is not demonstrated.

Take two flashlights.
I claim that the energy for the light produced by each
originates in the battery in each flashlight.
Is your claim otherwise?

Connect the negative terminals of the batteries together.
I claim that the energy for the light produced by each
still originates in the battery in each flashlight.
Is your claim otherwise?

Connect the positive terminals of the batteries together
through an ammeter. For the purposes of this experiment,
the batteries have exactly the same voltage so the ammeter
indicates that no current ever flows between the batteries.
I claim that the energy for the light produced by each
still originates in the battery in each flashlight.
Is your claim otherwise?

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

GO GUYS GO!

over 700 messages in this thread now, it would probalby be over 1000 already
if it hadn't had some contributions sucked away by the 'standing morphing
power ranger waves and other stupid stuff' thread!

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 12, 2:08*am, Roger Sparks wrote:
On Fri, 11 Jan 2008 18:49:20 -0800 (PST)

Keith Dysart wrote:
On Jan 11, 7:30*pm, Roger Sparks wrote:
On Fri, 11 Jan 2008 13:59:47 -0800 (PST)

Keith Dysart wrote:
On Jan 11, 11:15*am, Roger Sparks wrote:
On Thu, 10 Jan 2008 18:52:28 -0800 (PST)Keith Dysart wrote:
P(t) = V(t) * I(t)
by substitution from V = I * R
P(t) = V(t) * V(t) / R
and
P(t) = I(t) * I(t) * R

Correct.

But recall that power is the flow of energy per unit time.

The energy present in any section of line is the energy
present in the electric field of the capacitance plus
the energy present in the magnetic field of the inductance.

When the capacitance is in phase with the magnetic field, both are measured to peak at the same time. *This "resistive" condition is found in a transmission line when there are no standing waves. *Under resistive conditions, the electric field and inductive field merge into one electromagnetic field that carries only one energy. *The identical charges carry both electric and magnetic fields, with the fields in a ratio defined as impedance. Each field is a separate way of looking at the same energy.

I am not quite sure what you are attempting to say here and
whether you are disagreeing with my assertion that the total
energy in a section of line is equal to the sum of the energy
present in the capacitance and the energy present in the
inductance.

So from your expressions above
U = (CV^2)/2 + (LI^2)/2

You may have been reading about traveling waves. *They offer an explaination for how magnetic and electric fields may get out of phase. *When the two fields are out of phase, the U = (CV^2)/2 + (LI^2)/2 calculation will ALWAYS sum to more energy than can be found in the resistive wave. *This occurs because more energy actually is resident on the line with reflections, due to waves traveling in two directions. *The reflection effectively doubles the length of the transmission line available for energy storage. *

This seems like a complicated way to think about things. So
complicated that I will
not comment on correctness.

If P = (v^2)/r = (i^2)r is correct, what is the precondition that MUST be true?
Answer: The i and v measurements must be across a pure resistance, without reactance.

This is why it is often better just to use the original
equation:
P(t) = V(t) * I(t)
for it is always true and requires no preconditions.

I learned long ago that it DOES have a precondition. *The power measured will only be resistive if measured across a pure resistance. *If the measurement is across resistance plus reactance, then reactive power is also measured so P(t) = V(t) * I(t) becomes the sum of an unknown mix of two or more waves. * *

This too seems like a complicated way to think about things.

P(t) = V(t) * I(t) is simple, and always holds. One does not need to
consider
the wave content in any way. Or impedances. Measure the instaneous
voltage and
the instanteous current; multiplying will always yield the
instantaneous power.
Integrating instantaneous power will always yield the net energy
transfer over
the interval of integration.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 12, 2:08 am, Roger Sparks wrote:
I learned long ago that it DOES have a precondition. The power measured will only be resistive if measured across a pure resistance. If the measurement is across resistance plus reactance, then reactive power is also measured so P(t) = V(t) * I(t) becomes the sum of an unknown mix of two or more waves.

In power engineering, the power can consist of two components, active
power, V*I*cos(A) = watts, and reactive power, V*I*sin(A) = VARS.
These terms are defined in The IEEE Dictionary. P=V*I results in
Volt*Amperes, part active and part reactive if the voltage and current
are not in phase.
--
73, Cecil, w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 12, 8:10 am, Keith Dysart wrote:
P(t) = V(t) * I(t) is simple, and always holds.

But one must remember the units of that result is Volt*Amps, not
watts.
--
73, Cecil, w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

On Sat, 12 Jan 2008 03:49:18 -0800 (PST), Keith Dysart
wrote:

Take two flashlights.
I claim that the energy for the light produced by each
originates in the battery in each flashlight.
Is your claim otherwise?

Connect the negative terminals of the batteries together.
I claim that the energy for the light produced by each
still originates in the battery in each flashlight.
Is your claim otherwise?

Connect the positive terminals of the batteries together
through an ammeter. For the purposes of this experiment,
the batteries have exactly the same voltage so the ammeter
indicates that no current ever flows between the batteries.
I claim that the energy for the light produced by each
still originates in the battery in each flashlight.
Is your claim otherwise?

Hi Keith,

You have admitted that energy dissipated in the loads cannot be
identified to its source and you have used the above devices previous
to that statement. Your admission dismisses their applicability.
Separability is not demonstrated.

73's
Richard Clark, KB7QHC

Standing-Wave Current vs Traveling-Wave Current
 

On Sat, 12 Jan 2008 12:32:35 GMT, "Dave" wrote:

over 700 messages in this thread now, it would probalby be over 1000 already
if it hadn't had some contributions sucked away by the 'standing morphing

Chip,

This is not a terribly different posting from your own complaint.

73's
Richard Clark, KB7QHC

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 12, 11:13*am, Cecil Moore wrote:
On Jan 12, 8:10 am, Keith Dysart wrote:

P(t) = V(t) * I(t) is simple, and always holds.

But one must remember the units of that result is Volt*Amps, not
watts.

You really should put a bit of effort into understanding
the different expressions for power.

P(t) = V(t) * I(t)

is saying that the function which describes power with
respect to time, P(t), is the product of the function
describing voltage with respect to time, V(t), and the
function describing current with respect to time, I(t).

As corollary, the power at an instant of time is equal
to the voltage at that instant times the current at
that instant.

And, of course the unit watts is the same as the units
volts times amperes. Just do the substitution.

V = J/C
A = C/s

V A = (J/C) x (C/s)
= J/s
= W

You are confused with the common use in power
engineering where VA is used to mean Vrms*Irms.
Whether VA is equal to watts depends on the
phase relationship of the voltage and current.

When the voltage and current are not in phase, then
the expression
Pavg = Vrms * Irms * cos(A)
can be used to derive the average power.

This is a special case of P(t) = V(t) * I(t).
It is applicable when V(t) is a sinusoid of
a single frequency, and I(t) is related to V(t)
using the expression V(t) = Z * I(t).

Pavg = Vrms * Irms * cos(A)
is completely derived from P(t) = V(t) * I(t)
by appropriately substituting for V(t) and I(t)
and then averaging P(t) over an appropriate
interval (typically one cycle).

Bottom line:
P(t) = V(t) * I(t)
is always correct.

The other variations only work when
applied under the appropriate conditions.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 12, 1:22 pm, Keith Dysart wrote:
P(t) = V(t) * I(t)
As corollary, the power at an instant of time is equal
to the voltage at that instant times the current at
that instant.

You obviously mean the *real* instantaneous voltage times the *real*
instantaneous current. You should say that to avoid confusion.

Seems you need to add 'Re' to that equation to remove the obvious
ambiguity. P(t) = Re[V(t)]*Re[I(t)]

From "Fields and Waves", by Ramo & Whinnery:

W(t) = {Re[Vm*e^j(wt+A1)]}{Re[Im*e^j(wt+A2)]}
--
73, Cecil, w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 12, 2:05*pm, Cecil Moore wrote:
On Jan 12, 1:22 pm, Keith Dysart wrote:

P(t) = V(t) * I(t)
As corollary, the power at an instant of time is equal
to the voltage at that instant times the current at
that instant.

You obviously mean the *real* instantaneous voltage times the *real*
instantaneous current. You should say that to avoid confusion.

Seems you need to add 'Re' to that equation to remove the obvious
ambiguity. P(t) = Re[V(t)]*Re[I(t)]

From "Fields and Waves", by Ramo & Whinnery:

W(t) = {Re[Vm*e^j(wt+A1)]}{Re[Im*e^j(wt+A2)]}
--
73, Cecil, w5dxp.com

Good recovery.

But, of course, V(t) and I(t) are general functions
of time. In particular, the discusion was regarding
pulses.

Only when one constrains oneself to sinusoids and
chooses to use the complex exponential notation
is it necessary to use Re[].

But it was a good recovery.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 12, 3:25 pm, Keith Dysart wrote:
Only when one constrains oneself to sinusoids and
chooses to use the complex exponential notation
is it necessary to use Re[].

I think that's a false statement. Do you have any proof? Why would the
imaginary part of a pulse produce any power?
--
73, Cecil, w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 12, 3:25 pm, Keith Dysart wrote:
But, of course, V(t) and I(t) are general functions
of time. In particular, the discusion was regarding
pulses.

Pulses can be analyzed as Fourier sinusoidal functions with multiple
frequencies. Point is that you could have saved a month of grief on
this newsgroup if you had initially said your power equation applied
only to real voltage and real current. If you had done that, nobody
would have argued with you.
--
73, Cecil, w5dxp.com