Standing-Wave Current vs Traveling-Wave Current
 

Dave Heil wrote:
Sorry, Cecil, it isn't an illusion. Sunrise and sunset are simply names
we've given to that which you've described. They really take place and
are not at all an illusion.

You can believe that the sun rises and sets if you want to,
Dave. Most of us know that the sun is fixed at the center
of the solar system. The names "sun rise" and "sun set" were
coined back when humans believed that the sun actually made
a trip across the sky every day over a flat earth. Do you also
believe in a flat earth?
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:
Cecil Moore wrote:
Roy Lewallen wrote:

The existence of both voltage and current at any point along the line
tells us that there is instantaneous power at that point, ...

Not if the voltage and current are always 90 degrees out of
phase which is a fact of physics for pure standing waves. There
is no power, instantaneous or otherwise, in pure standing waves.
The cosine of 90 degrees is *always* zero.

These comments in total are very interesting, by both authors. Thank
you for them.

It is clear that there is not a standard way of describing energy that
is in the process of being stored in either an inductor or capacitor.
Clearly it is stored, not used (converted) as in a resistor.

Sure there is. The energy stored in a capacitor is 1/2 * C * V^2, and in
an inductor, 1/2 * L * I^2. The rate of energy flow into or out of an
inductor or capacitor is called "power". It's exactly the same stuff as
energy moving into or through a resistor, and it's measured and
described in exactly the same way.

Storage of energy in a capacitor (for instance) occurs over time and
requires power to complete.

Yes, power is the rate of energy transfer.

So how do we describe that power if it is
always zero because voltage and current are 90 degrees out of phase?

See the following example which shows just that situation.

Should we recognize that only the peak voltage and peak current is 90
degrees out of phase, with the entire charging time occurring within
those two time extremes? Between those two time extremes, the voltage
and current are in phase but at changing impedance, with power flowing
into the capacitor. So think I.

The whole concept of impedance gets a bit flaky when working in the time
domain. I recommend sticking to voltage, current, power, and energy when
doing so. Go through the following example, and I believe you'll get a
much better idea of what's happening on a basic level.

But you've made some good observations. The answer is that, as I
mentioned, the instantaneous power (p(t)) is not zero except at the ends
and middle of the example line. Some of the posters are confusing
average and instantaneous power, which is leading to incorrect
statements and conclusions. Instantaneous power can easily be non-zero
while maintaining zero average power, as I'll illustrate.

Let's look, for example, at the half wavelength open circuited line
described by "Dave", being driven by v = 0.5 sin(wt), 1/8 wavelength
from the end. At that point,

vf(t) = 0.5 sin(wt - 135 deg.)
vr(t) = 0.5 sin(wt - 225 deg.)
if(t) = 0.01 sin(wt - 135 deg.)
ir(t) = -0.01 sin(wt - 225 deg.)

considering the positive direction of ir(t) as the same as the positive
direction of if(t) so we can add if(t) and ir(t) to get i(t).

Using elementary trig identities for the addition, and dropping the
explicit identification of degrees for simplicity:

vf(t) + vr(t) = v(t) = sin(wt - 180) * cos(45) = 0.7172 * sin(wt - 180)
if(t) + ir(t) = i(t) = 0.02 * cos(wt - 180) * sin(45) = 0.01414 *
cos(wt - 180)

One important thing to notice is that, unlike the ends or center of the
line, neither the voltage nor the current is zero at this point.
Consequently, their product is also non-zero.

The voltage and current are, of course, 90 degrees out of phase, as they
are everywhere along an open circuited or short circuited line, or one
terminated in a pure reactance. And when the average power is computed
from the instantaneous power, the result will be zero. But that doesn't
mean that the instantaneous power is zero as some are claiming. Let's
see what it does mean. Using another trig identity,

p(t) = v(t) * i(t) = 0.01 * sin(wt - 180) * cos(wt - 180)
= 0.005 * sin(2wt)

This is the instantaneous power which, as I've described before, is a
sinusoidal function with rotational frequency 2w. It has no offset, so
the average value is zero. But it shows that energy moves back and forth
past this point every cycle. Equal amounts move each direction each
cycle, so no net power flows either direction.

There are simpler ways to get the same answer, but this one shows very
basically where the energy is going at every instant. If instead you
start out by throwing away the time information and just looking at
averages, you lose a lot of information along with the basic
understanding of energy movement in the system. I'm afraid some people
have taken this simpler approach without realizing what information and
insight they've lost by doing so. The average value can always be
determined from the instantaneous function, but never the other way around.

I think of the "standing wave" as being equivalent to the graph on my
power bill that shows power used daily over a month. Granted that the
standing wave is recorded in V or I, but we only need to know the R that
the standing wave is acting through to determine the power that it
represents.

Not quite. The power bill graph is a graph of the power, much like the
p(t) function I calculated. The energy, which you pay for, is the
integral of that graph. If you used a constant power of 1 kW for half
the month, then sent 1 kW back into the power grid for the other half of
the month, you'd technically owe nothing, since you'd use no net power.
You might even have stored it somewhere and given the very same energy
back rather than using some and generating an equal amount later. That's
exactly what's happening at the point on the line 1/8 wavelength from
the end.

Roy Lewallen, W7EL

-- I've had many opportunities when doing this to make arithmetic or
trigonometric errors, and I haven't taken the time to thoroughly check
my work -- although the conclusion is correct. I'd appreciate anyone
pointing out any mathematical errors he's found.

Standing-Wave Current vs Traveling-Wave Current
 

Not if the voltage and current are always 90 degrees out of
phase which is a fact of physics for pure standing waves. There
is no power, instantaneous or otherwise, in pure standing waves.
The cosine of 90 degrees is *always* zero.
--
************************************************** *******************************8

Cecil, your technical and mathematical skills vastly exceed mine... My
poor little monkey brain has to work in the concrete, not the
abstract...
I sort of understand the concept of the standing wave being the
instantaneous vector product of two colliding EM wave fronts...
I accept the fact that MATHEMATICALLY; P = V*I* cos (theta).... And
being that theta is DEFINED as 90 degrees the cos = 0, so the
instantaneous power is zero...
But here is where the math breaks down...
On my herd of 100KW RF generators... Every time the load failed for
even a split second, those fire breathing dragons would blow a hole
through the quarter inch thick slabs of copper that made up the line,
instantaneously...
To prove that mathematically there is no power contained in a standing
wave collides with my, farm boy, real world, physics experiments...
The voltage peak on that line when that standing wave forms is real,
it has mucho power (esp when driven by the dragon breath of a 100KW
triode with 15KV on the plate), and it blows the hole in the line at
the same distance from the load every time (proof of standing wave
theory)...

This is why i have a problem accepting that a standing wave contains
no power - regardless of the prominence of the men who wrote the
textbooks..

cheers ... denny

Standing-Wave Current vs Traveling-Wave Current
 

"Denny" wrote in message
...
Not if the voltage and current are always 90 degrees out of
phase which is a fact of physics for pure standing waves. There
is no power, instantaneous or otherwise, in pure standing waves.
The cosine of 90 degrees is *always* zero.
--
************************************************** *******************************8

Cecil, your technical and mathematical skills vastly exceed mine... My
poor little monkey brain has to work in the concrete, not the
abstract...
I sort of understand the concept of the standing wave being the
instantaneous vector product of two colliding EM wave fronts...
I accept the fact that MATHEMATICALLY; P = V*I* cos (theta).... And
being that theta is DEFINED as 90 degrees the cos = 0, so the
instantaneous power is zero...
But here is where the math breaks down...
On my herd of 100KW RF generators... Every time the load failed for
even a split second, those fire breathing dragons would blow a hole
through the quarter inch thick slabs of copper that made up the line,
instantaneously...
To prove that mathematically there is no power contained in a standing
wave collides with my, farm boy, real world, physics experiments...
The voltage peak on that line when that standing wave forms is real,
it has mucho power (esp when driven by the dragon breath of a 100KW
triode with 15KV on the plate), and it blows the hole in the line at
the same distance from the load every time (proof of standing wave
theory)...

This is why i have a problem accepting that a standing wave contains
no power - regardless of the prominence of the men who wrote the
textbooks..

cheers ... denny

Same here, I burn the coil with "no power" on standing wave circuit -
quarter wave resonant loaded whip.
Cecil is telling us that the current becomes automatically traveling wave,
while EZNEC model shows drop along the coil and rest of the circuit with
nice standing wave.
I grew up in the city, around cars and I am having problem swallowing that
we have SW current and voltage but no power, while we pump power into the
circuit and see nasty things happening requiring power.

three cheers... Yuri oK3BU

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:
Dave Heil wrote:
Sorry, Cecil, it isn't an illusion. Sunrise and sunset are simply
names we've given to that which you've described. They really take
place and are not at all an illusion.

You can believe that the sun rises and sets if you want to,
Dave. Most of us know that the sun is fixed at the center
of the solar system. The names "sun rise" and "sun set" were
coined back when humans believed that the sun actually made
a trip across the sky every day over a flat earth.

Sunrise and sunset are simply names we've given to that which you've
described. My logging program provides me with sunrise and sunset times
for various parts of the world. The National Weather Service uses those
terms. You may choose to say at the beginning of each day that the
earth has rotated so that the star at the center of our solar system is
visible. I don't know anyone who spouts that kind of mouthful.

Do you also
believe in a flat earth?

Where are you getting the "also"?

Now, what about the magic tricks?

Dave K8MN

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
The voltage and current are, of course, 90 degrees out of phase, ...

You have just contradicted yourself. There *cannot* be any
real power associated with a voltage and current that are
90 degrees out of phase.

Power = V*I*cos(90) = 0 watts

I will attempt to check your math. I'll bet your dimensions
are volt*amps and NOT watts.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Denny wrote:
But here is where the math breaks down...
On my herd of 100KW RF generators... Every time the load failed for
even a split second, those fire breathing dragons would blow a hole
through the quarter inch thick slabs of copper that made up the line,
instantaneously...

The math didn't break down, Denny, your logic did.

1. The standing wave voltage is *ALWAYS* 90 degrees out of
phase with the standing wave current, by definition.

2. It is *IMPOSSIBLE* to "blow a hole through the quarter
inch thick slabs of copper" with a voltage and current that
are 90 degrees out of phase.

3. Therefore, the voltage and current that blew the holes
in the copper slabs were NOT 90 degrees out of phase and
therefore were not standing waves. In-phase voltage and
current is required to blow holes in copper slabs.

4. Standing waves are a steady-state phenomenon. Blowing
holes in copper slabs is a transient phenomenon. At the
end of steady-state, the standing wave energy is always
converted into real watts and either radiated or dissipated.

The energy that blew holes in your copper slabs did NOT
meet the definition of "standing wave energy". During
the transient state, it was converted from standing wave
energy (with its 90 degree phase difference between voltage
and current) to traveling wave energy (with its in-phase
relationship between voltage and current).
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Yuri Blanarovich wrote:
Cecil is telling us that the current becomes automatically traveling wave,

Yes, it is impossible to get any watts out of a voltage
and current that are 90 degrees apart simply because
there are zero watts available.

Standing wave voltage and current are 90 degrees apart,
by definition.

Therefore, by definition, any wave that yields watts
is NOT a standing wave.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
The voltage and current are, of course, 90 degrees out of phase, ...

p(t) = v(t) * i(t) = 0.01 * sin(wt - 180) * cos(wt - 180)
= 0.005 * sin(2wt)

Those are not watts, those are volt*amps. Therefore that's
not real power. Real power is:

p(t) = v(t) * i(t) * cos(90) = ZERO
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Dave Heil wrote:
Sunrise and sunset are simply names we've given to that which you've
described.

Those words were coined when the human race believed
that the sun actually traveled across the sky each
day. The sun "rising" means the sun is moving but
that is just an illusion.

You seem to believe that illusions don't exist in
reality. That very belief is an illusion.
--
73, Cecil http://www.w5dxp.com