Michael Coslo wrote:
Roy Lewallen wrote:


My postulate is that Newton was wrong: moving objects come to a rest
without any external applied force. Every observation made supports
this. There's no need to consider what happens in a frictionless
environment, since such a thing doesn't exist.


Isn't that like lossless wires, perfect grounds, and other such?

The conditions that cause an object to slow and stop in real life
are the proof of the law. To the contrary, it proves Newton correct. The
forces act just as they should.

- 73 de Mike N3LI -

To repeat my posting of Dec. 8:

--------------

This would be funny if it weren't sad. This newsgroup is one of the few
places I can think of where the silly statement I posted about moving
bodies and friction would be taken seriously. But it's really no
surprise, since it's much less unreasonable than the imaginative
alternative theories which are seriously presented, and just as
seriously argued, here daily.

My postulate about objects in motion was a parody of Cecil's rejection
of theoretical cases on the basis that they can't exist in practice, my
intent being to show how such a rejection leads to incorrect results.
But I see it's drawing the same serious response as Cecil's and Art's
postings. All that's missing is one of Richard's quotes from Terman and
support from Derek.

--------------

Roy Lewallen, W7EL

Roy Lewallen wrote:
My postulate about objects in motion was a parody of Cecil's rejection
of theoretical cases on the basis that they can't exist in practice,

Roy, once again you distort what I have said.
I did NOT reject any theoretical cases. I said
I personally don't have time to consider those
cases as my daughter is facing emergency surgery
in New York state and I am standing by to hop a
plane during the Christmas season rush.

If you know more about my personal time than I do,
please let me know exactly how you accomplish that
feat. If your argument is that gurus know everything,
I will certainly understand. You have used that
argument before.
--
73, Cecil http://www.w5dxp.com

Roy Lewallen wrote:
This would be funny if it weren't sad.

I'll tell you what would be funny if it weren't
sad. You present yourself as some omniscient guru
on this newsgroup yet you recently showed your
absolute ignorance of standing waves and the math
behind that subject.

Most of the knowledgeable people on this newsgroup
now know that, as far as standing waves go, you are
just a flimflam man. I am amazed that you have the
balls to post anything else after that fiasco.
--
73, Cecil http://www.w5dxp.com

Keith Dysart wrote:

On Dec 9, 9:36 pm, Roger wrote:



The constantly-in-phase traveling wave concept requires the difficult-to-believe observation that a directional ammeter placed very near the end of an open transmission line will read the same current as if it were placed at the source end. Perhaps someone can perform that experiment some day, but I can not imagine how it can be done without placing a load on the line, thus invalidating the initial assumptions.



The experiment will show the expected result but will not help understand why. For that, examination of the measurements and arithmetic performed by a directional ammeter is useful. Below, all voltages and currents are instantaneous. Total voltage, Vt = Vf + Vr Total current, It = If - Ir Vf = If * Z0 Vr = Ir * Z0 Substituting.... Vt = (If + Ir) * Z0 Ir = Vt/Z0 - If If = It + Ir If = It + (Vt/Z0 - If) If = (It + Vt/Z0)/2 Similarly, Ir = (It - Vt/Z0)/2 The directional ammeter measures instantaneous Vt and It, does the above arithmetic and presents If. A directional ammeter that presents a single number rather than the time varying If has probably converted the instantaneous values to RMS. Examing It and Vt at various points on the line and doing the above arithmetic will reveal why the same value for If is obtained everywhere. Directional wattmeters are more common than directional ammeters. A directional wattmeter does the above arithmetic then squares If, multiplies by Z0 and presents the results in watts. All this from just measuring Vt and It. ....Keith

Hi Keith,

Thanks to you and others for responding on this side issue.   It was very helpful to me and resulted in a vast improvement in how I understood the theory behind directional watt meters.  I had the misconception that current pickup over some lineal distance of transmission line was NECESSARY for the device to work, but now clearly understand that instantaneous measurement points suffice (and that instantaneous current measurement may be impossible).

After considerable thought, I think the math you presented above is for one of two cases of reflective waves, the reflection from a higher  impedance load.   When the load is less than the Zo of the line, the currents add but voltages subtract.   Right?  The end result is the same for both cases.

To elaborate,  the power (Pt) resident on the transmission line will always be the sum of forward power and reflected power.  Mathematically:

Ptotal = Pf + |Pr|
Pt = If * Vf  +  |(-Ir * Vr)|   (Case of load impedance greater than line impedance) (Case 1)
or
Pt =  If * Vf  +  |(Ir  *  - Vr)|  (Case of load impedance less than line impedance) (Case 2)

Notice that Pt is always the ABSOLUTE SUM of the two power terms because the reflective power always carries a negative sign on either the current or voltage term.  As a result, the summed  value of the resident power on a mismatched line is always greater than the actual power going to the load.  

Your example was correct for Case 1.   I will show only Case 2.

Vt = Vf - Vr
It = If + I r

Vf = If * Z0
Vr = Ir * Z0

Substituting...

Vt = (If - Ir) * Z0
Ir = If - Vt / Z0

If = It - Ir
If = It - If + Vt/Z0
If = (It + Vt/Z0)/2

Similarly, Ir = (It - Vt/Z0)/2

The results are the same for both Case 1 and Case 2.

Agreed?

73, Roger, W7WKB

Roger wrote:
To elaborate, the power (Pt) resident on the transmission line will always be
the sum of forward power and reflected power.

That statement is slightly misleading. The length of the transmission
line can be converted to time, e.g. one microsecond long. If the
forward power is 100 joules/sec and the reflected power is 50
joules/sec, the total energy in the one microsecond long transmission
line will be 100+50 = 150 microjoules. Energy is what is flowing in
the feedline and must be conserved, not power.

The technically correct way to say what you are trying to say is:
There is exactly the amount of energy resident in the transmission
line needed to support the steady-state forward and reflected power
readings. I realize that I am picking nits.
--
73, Cecil http://www.w5dxp.com

On Dec 11, 4:21 pm, Cecil Moore wrote:
Jim Kelley wrote:
Yes, I'm very familiar with that article. You've already posted a link
to it dozens of times on this newsgroup. It very clearly illustrates
exactly those thing which I may have somewhat more 'colorfully' restated
above, and more. It includes equations with variables for forward and
reflected power all throughout,

Yes, forward and reflected power measured at a *FIXED* measurement
point. There is no "power flow" anywhere in my article. Energy
does the flowing. Power is the measurement at a *FIXED* measurement
point of that energy flow past that *FIXED* point. Did you note
the use of the word, "FIXED"?

Even though all my references, including the IEEE Dictionary
allow for "power flow", I avoided it in my article as a favor
to you.

a reference to a supposed "4th mechanism
of reflection" (that's the magical way in which waves of power and
energy change direction),

Yes, that may be somewhat original and therefore frightening
for you. Galileo would have scared you to death. If, as Walter
C. Johnson says, interference can *cause* standing waves, it
can probably also cause reflections at an impedance discontinuity
through wave cancellation. You are going to have to do more
than wave your hands to prove otherwise. Not refusing to
answer my questions about my examples would be a good start.

Remember your absolute refusal to compute the total joules/sec
after the first internal reflection arrived at the thin-film
anti-reflective coating in my example? When you learn how to
properly manage irradiance, get back to us.

and illustrations with arrows named Pref
showing how power is reflected at impedance discontinuities.

No! No! No! Power is NOT reflected at an impedance discontinuity.
Those are Poynting vectors. Energy is reflected and that reflected
energy is measured and called "reflected power". Reflected power
is not moving. You continue to get it wrong. The reflected power
is acutally reflected energy measured flowing past a *FIXED* point
near the impedance discontinuity. There are joules in the reflected
wave. The joules in the reflected wave are measured flowing past
a *FIXED* measurement point. But, of course, I have explained
all of this to you before yet you continue bear false witness
after all these years.

Back when our corresponence was more cordial, I advised you not to write
those things. And now you'd like to deny having done it; all the while
portraying me as a liar. You're beautiful, man.

I changed my article just to make you happy. You obviously
have misunderstood, either through lack of processing power,
ignorance, or deliberately. I would guess it is deliberate.

The debate has never been aoout a little looseness
in the terminology; i.e. does "power flow" or does
"energy flow". That is a straw man of your own
making.

The debate is over much more fundamental issues.

....Keith

On Dec 11, 10:34 pm, Roger wrote:
Keith Dysart wrote:On Dec 9, 9:36 pm, wrote:The constantly-in-phase traveling wave concept requires the difficult-to-believe observation that a directional ammeter placed very near the end of an open transmission line will read the same current as if it were placed at the source end. Perhaps someone can perform that experiment some day, but I can not imagine how it can be done without placing a load on the line, thus invalidating the initial assumptions.The experiment will show the expected result but will not help understand why. For that, examination of the measurements and arithmetic performed by a directional ammeter is useful. Below, all voltages and currents are instantaneous. Total voltage, Vt = Vf + Vr Total current, It = If - Ir Vf = If * Z0 Vr = Ir * Z0 Substituting.... Vt = (If + Ir) * Z0 Ir = Vt/Z0 - If If = It + Ir If = It + (Vt/Z0 - If) If = (It + Vt/Z0)/2 Similarly, Ir = (It - Vt/Z0)/2 The directional ammeter measures instantaneous Vt and It, does the above arithmetic and presents If. A directional ammeter that presents a single number rather than the time varying If has probably converted the instantaneous values to RMS. Examing It and Vt at various points on the line and doing the above arithmetic will reveal why the same value for If is obtained everywhere. Directional wattmeters are more common than directional ammeters. A directional wattmeter does the above arithmetic then squares If, multiplies by Z0 and presents the results in watts. All this from just measuring Vt and It. ...KeithHi Keith,
Thanks to you and others for responding on this side issue. It was very helpful to me and resulted in a vast improvement in how I understood the theory behind directional watt meters. I had the misconception that current pickup over some lineal distance of transmission line was NECESSARY for the device to work, but now clearly understand that instantaneous measurement points suffice (and that instantaneous current measurement may be impossible).
After considerable thought, I think the math you presented above is for one of two cases of reflective waves, the reflection from a higher impedance load. When the load is less than the Zo of the line, the currents add but voltages subtract. Right?

I don't think so.
Vt = Vf + Vr, It = If - Ir, Vf = If * Z0 and Vr = Ir * Z0 are the
fundamental
equations defining forward and reverse waves.

Perhaps you arrive at two choices because sometimes Vr and Ir
are negative, which after simplification appears to give an
alternate form?

The end result is the same for both cases.

This is good. If you chase the signs, though, I think you will find
that there is only one case.

We probably should not toss Power into the mix until agreement
is reached on this. Power is fraught with issues which seriously
confuse some.

...Keith

Keith Dysart wrote:
The debate has never been about a little looseness
in the terminology;

Actually, I believe that every disagreement between
Jim Kelley and myself has been semantic in nature.
There are no technical fundamentals upon which we
disagree.

We both agree that a light wave from Alpha Centauri
that hits the earth has transferred energy.

We disagree about a light wave from Alpha Centauri
that misses the earth. I say the energy in the light
wave is in the process of being transferred. Jim
disagrees.

The debate is over much more fundamental issues.

Like what? The definition of "transfer"?
--
73, Cecil http://www.w5dxp.com

Keith Dysart wrote:
On Dec 11, 10:34 pm, Roger wrote:
When the load is less than the Zo of the line, the currents
add but voltages subtract. Right?

I don't think so.
Vt = Vf + Vr, It = If - Ir, Vf = If * Z0 and Vr = Ir * Z0 are the
fundamental equations defining forward and reverse waves.

I think Roger is talking about the phase shift at a reflection
point. If ZL Z0, the current reverses phase, i.e.
It = |If| - |Ir| = 0. If ZL Z0, the voltage reverses phase,
i.e. Vt = |Vf| - |Vf| = 0

The sign on the reflected current is just a directional
convention left over from DC and is unnecessary as the phase
angle of the phasor values takes care of the signs.
Interestingly, the field of optics has a different convention.
--
73, Cecil http://www.w5dxp.com

Keith Dysart wrote:
The debate is over much more fundamental issues.

I guess that depends upon the definition of
"fundamental" doesn't it? :-) Hint: Virtually
every verbal disagreement is semantic.
--
73, Cecil http://www.w5dxp.com