On Tue, 26 Aug 2003 18:29:59 -0400, wrote:


It would seem that if we are allowed to posit, for the purposes
of discussion, the existence of ideal transmission lines, we should
also be allowed to posit the existence of ideal observation tools.


Hi Keith,

All fine and well, but you have simply "defined" your own solution.
You haven't explored it, explained it, or learned from it. Saying "it
is thus" is meaningless. It's like Cecil's logic of zero power
arising from the combination of all possible powers. "A gazzillion
watts facing -gazzillion watts is zero watts - how can you deny a
gazzillion watts?"

You may allow yourself these ideal tools, but you have wholly
neglected their description, method, and application. The same
solutions exist in reality, they are simply rough around the edges.
Knowing the scope of error in context let's us judge for ourselves how
significant the argument is. As I pointed out, there are a number of
metaphors available to do this, and they still go wanting.

This so-called "ideal" is an illusion. I've measured physical
constants out so many decimal places that the random noise of the big
bang limited its accuracy, and yet there were still more places to
report if anyone wished hard enough to believed in it. Perfection
allows for error too.

73's
Richard Clark, KB7QHC

W5DXP wrote:

wrote:
W5DXP wrote:
Actually, not. You continue to resist pointing out which step is wrong.

You obviously have not read all of my replies.

On the contrary. But finally this one actually addresses
one of the steps rather than just attempting to show that that
conclusion is wrong.

Is it step 2)?
"2) At quarter wave points along this line, voltages and currents
which are always 0 can be observed -- the standing wave"

This seems to be generally accepted.

This is where your confusion starts. There are points where the
NET voltage and NET current are zero. Those are merely the points
where the forward Poynting vector and reflected Poynting vector
are out of phase.

This emphasis on NET seems to be the place where the difficulties
begin. An (ideal) voltmeter placed at a voltage minima on the line
indicates a voltage of 0 volts. You appear to be saying that
despite this indication of 0, there is actually voltage present.

This contention leads to a number of questions related to voltmeters:
- when I use a voltmeter to measure the voltage in a circuit,
how do I know when an indication of 0 means 0?
- how do I know when an indication of 0 means there are
really a number of voltages which sum to 0?
- how do I determine what these voltages are which sum to 0?
- if it indicates other than 0, is it really indicating a
number of voltages which sum to the result?
- how do I determine what these voltages actually are?
- can voltmeters be trusted at all?
- when?

In my world, voltmeters indicate volts. There is no need for
second guessing.

Until this very fundamental difference is settled, there is
no value in examining the remaining steps.

....Keith

On Tue, 26 Aug 2003 21:45:36 -0400, wrote:


This emphasis on NET seems to be the place where the difficulties
begin. An (ideal) voltmeter placed at a voltage minima on the line
indicates a voltage of 0 volts. You appear to be saying that
despite this indication of 0, there is actually voltage present.


Hi Keith,

We covered this before. The NET 0 "could be" the result of a
bajillion volts and -bajillion volts. Now you couldn't possibly
ignore a bajillion volts now, could you? ALL things are possible when
we have a zero to rummage up arguments that revolve around what "could
be." It takes only a slight rhetorical slide to become "is."

You certainly would need a perfect voltmeter to withstand the
potential fields nearby. ;-)

73's
Richard Clark, KB7QHC

wrote:
This emphasis on NET seems to be the place where the difficulties
begin. An (ideal) voltmeter placed at a voltage minima on the line
indicates a voltage of 0 volts. You appear to be saying that
despite this indication of 0, there is actually voltage present.

There are actually two voltages present of equal amplitude and
opposite phase. Their phasor sum is zero volts.

- how do I determine what these voltages actually are?

|Vfwd| = Sqrt(Pfwd*Z0) |Vref| = Sqrt(Pref*Z0)

If the voltmeter reads zero, these two voltages are equal in magnitude
and opposite in phase. Do you know how to sum two phasor voltages?
--
73, Cecil http://www.qsl.net/w5dxp



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Hm, this has me puzzled. Assuming a purely real Z0, you're taking the
square roots of two purely real quantities, each of which can have two
values, and speaking of a phase angle between them. Where in the process
did they pick up phase information? Or do you just mean when one is the
negative of the other? If so, how do you tell -- each has two roots,
that is, Sqrt(Pfwd*Z0) can be either positive or negative, and so can
Sqrt(Pfref*Z0). How do you know when one is the negative of the other?

Roy Lewallen, W7EL

W5DXP wrote:
Richard Clark wrote:

We covered this before. The NET 0 "could be" the result of a
bajillion volts and -bajillion volts.


It could be but it's not. The NET 0 is the result of Pfwd=Pref
Thus, Sqrt(Pfwd*Z0) = Sqrt(Pref*Z0) and when they are 180 deg
out of phase, the net voltage is zero.

Keith wrote:
"-how do I determine what these voltages actually are?"

You measure the power in either direction using a wattmeter containing a
directional coupler. The volts, amps, and power flowing each direction
are the same. That`s why you have zeros.

Then, P = Esquared / Zo, so E = sq rt (P)(Zo)

Best regards, Richard Harrison, KB5WZI

I wrote:
"E = sq rt (P)(Zo)"

Some may not realize that E is an rms value. Keith early in the thread
was inserting instantaneous volts which can be an infinite range of
values as they may be taken at any point in a cycle for evaluation. I
agree with Cecil`s "tits on a boar hog" characterization of value in
transmission line problem utility.

Best regards, Richard Harrison, KB5WZI

Roy Lewallen wrote:
Hm, this has me puzzled.

:-) Good one, Roy. :-)

Assuming a purely real Z0, you're taking the
square roots of two purely real quantities, each of which can have two
values, and speaking of a phase angle between them. Where in the process
did they pick up phase information? Or do you just mean when one is the
negative of the other? If so, how do you tell -- each has two roots,
that is, Sqrt(Pfwd*Z0) can be either positive or negative, and so can
Sqrt(Pfref*Z0). How do you know when one is the negative of the other?

V^2/Z0=P is a well known equation (so is I^2*Z0=P). These are *RMS* values.
So the RMS voltage is V = Sqrt(Pfwd*Z0). Root Mean Square AC voltages are
equivalent to DC voltages in power dissipation and are generally considered
to be positive values because they are the sum of squared terms. We can turn
those RMS voltages into phasors by adding the phase angles. When Vfwd+Vref = Vmax,
Vfwd and Vref are in phase (at the SWR voltage maximum point). When Vfwd+Vref = Vmin,
Vfwd and Vref are 180 degrees out of phase (at the SWR voltage minimum point).
Vmax/Vmin = VSWR. Please feel free to pull my leg again anytime. :-)
--
73, Cecil http://www.qsl.net/w5dxp



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Richard Clark wrote:
Certainly we cannot ignore the bajillion volts *RMS*, now, can we?
;-)

We can unless Sqrt(P*Z0) equals a bajillion volts which it usually
doesn't. These values are all inter-related. A 200W transmitter
will pour Sqrt(200*50) = 100 volts RMS into a 50 ohm load. That's
a no-brainer. Why do you act like it is a far out big deal?
--
73, Cecil http://www.qsl.net/w5dxp



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On Wed, 27 Aug 2003 15:51:20 -0500, W5DXP
wrote:

Why do you act like it is a far out big deal?

Hi Cecil,

Are you now trying to convince us that a bajillion volts *RMS* meeting
-bajillion volts *RMS* is NOT NET 0? Those are merely the points
where the forward Poynting vector and reflected Poynting vector
are out of phase. We are dealing with component energies. The NET
energy is zero. The component energies are not zero. This would seem
to conflict with much of your wave mechanics for the last 6 months.
Shirley you cannot deny the impact of a bajillion volts *RMS* simply
because your current argument doesn't need that solution, can you? (I
guess you can.)

However, that is not to say that we didn't notice that rhetorical
slide from "could be" to the firmer "is." We can specify these
bajillion volts *RMS* with a phase of a mega-bajillion degrees (with
deliberate care to avoid problematic 179 or 181 degrees) to suit any
opportunistic need. ;-)

73's
Richard Clark, KB7QHC