I really meesed up that last paragraph!

Dave wrote:

You can do the same with DC - you don't need to use AC at all. Put a 50
V battery in series with the pure 50 Ohm load and it supplies 50 W. Put
it in series with another load, consisting of a 50 Ohm voltage source in
consisting of a 50 volt voltage source, not a 50 Ohm voltage source!
series with a 50 Ohm load, and it is no surprise it delivers a different
power. Depending on what way you connect the two batteries, the current
would be 0 A or 4 A, and so the power 0 or 200W.
As I said before, 0 or 2A, which gives 0 or 200 W.

Dave wrote:

You can do the same with DC - you don't need to use AC at all. Put a 50 V
battery in series with the pure 50 Ohm load and it supplies 50 W. Put it in
series with another load, consisting of a 50 Ohm voltage source in
consisting of a 50 volt voltage source, not a 50 Ohm voltage source!
series with a 50 Ohm load, and it is no surprise it delivers a different
power. Depending on what way you connect the two batteries, the current would
be 0 A or 4 A, and so the power 0 or 200W.
As I said before, 0 or 2A, which gives 0 or 200 W.

Your example, though correct, has little to do with the case being discussed.

In your example it is true that power varies depending on what way you connect
the two batteries, but in all cases the total power dissipated in the loads
remains equal to the total power delivered by the sources.

In the case being discussed instead the power dissipated in the load varies with
NO CHANGE in total power delivered by the sources.

73

Tony I0JX

On Sat, 17 Nov 2007 22:05:57 +0100, "Antonio Vernucci"
wrote:

In the case being discussed instead the power dissipated in the load varies with
NO CHANGE in total power delivered by the sources.

Hi Tony,

Buying into a blighted argument (what Cecil presented) leads to some
very strange contortions such as you describe above. That, or there
are some strained language problems here.

73's
Richard Clark, KB7QHC

Richard Clark wrote:
"Antonio Vernucci" wrote:
In the case being discussed instead the power dissipated in the load varies with
NO CHANGE in total power delivered by the sources.

Buying into a blighted argument (what Cecil presented) leads to some
very strange contortions such as you describe above.

Sorry Richard, but if you listen to Antonio, you might learn
something about conservation of energy. He is one of the few
posters who seems to have a grasp of the technical facts that
have, so far, eluded you and other gurus.
--
73, Cecil http://www.w5dxp.com

Antonio Vernucci wrote:
Dave wrote:

You can do the same with DC - you don't need to use AC at all. Put a
50 V battery in series with the pure 50 Ohm load and it supplies 50
W. Put it in series with another load, consisting of a 50 Ohm voltage
source in
consisting of a 50 volt voltage source, not a 50 Ohm voltage source!
series with a 50 Ohm load, and it is no surprise it delivers a
different power. Depending on what way you connect the two batteries,
the current would be 0 A or 4 A, and so the power 0 or 200W.
As I said before, 0 or 2A, which gives 0 or 200 W.

Your example, though correct, has little to do with the case being
discussed.

In your example it is true that power varies depending on what way you
connect the two batteries, but in all cases the total power dissipated
in the loads remains equal to the total power delivered by the sources.

yes

In the case being discussed instead the power dissipated in the load
varies with NO CHANGE in total power delivered by the sources.

no way. Prove there is no change in the power delivered by the sources.

no way. Prove there is no change in the power delivered by the sources.

Making reference to the case under discussion (Wave#1 produces 50 joules/s at
the receiver when alone, and so does Wave#2 when alone), imagine that the two
waves are generated by two remote transmitters (+antennas) and that you measure
the power of the two superimposed waves on a receiver (+antenna) that you can
move in the space as you like.

If you put your receiver/antenna in a point where the two waves have equal
amplitude and opposite phase, your receiver will measure zero joules/s.

If you instead put your receiver/antenna in a point where the two waves have
equal amplitude and same phase, your receiver will measure 200 joules/s (i.e.
four times the power produced by each wave alone, not just two times).

Finally, if you put your receiver in a point where the two waves have equal
amplitude and a 45 deg. shift (as in the proposed case), your receiver will
measure 171 joules/s (still more than twice the power produced by each wave
alone).

Moving your receiver here and there will obviously cause no change in the power
delivered by the two remote transmitters.

The trick is due to the fact that the two waves interfere each other in
constructive or destructive manner depending on the particular receive point.
So, in the "lucky" points you get some extra power, which is however compensated
for by the power loss occurring at the "unlucky" points.

The original question is deceiving, because it attracts the reader's attention
on just one particular point of the space, where energy can unexplicably appear
to be created or destroyed. But instead considering the power distribution over
the whole space, the mistery disappears.

73

Tony I0JX

Forgot to also mention that, obviously, shutting off one of the two remote
transmitters causes no change in the power delivered by the other transmitter.

73

Tony I0JX

Antonio Vernucci wrote:
. . .
The original question is deceiving, because it attracts the reader's
attention on just one particular point of the space, where energy can
unexplicably appear to be created or destroyed. But instead considering
the power distribution over the whole space, the mistery disappears.

Exactly. Misdirection is the primary tool used by magicians
(illusionists) to distract us into thinking something is occurring which
really isn't. Its utility hasn't been lost on those wanting to divert
our attention from the flaws in their arguments.

Pay no attention to the man behind the curtain. . .

Roy Lewallen, W7EL

Exactly. Misdirection is the primary tool used by magicians (illusionists) to
distract us into thinking something is occurring which really isn't. Its
utility hasn't been lost on those wanting to divert our attention from the
flaws in their arguments.

Anyway, re-thinking on the case being here discussed, we must admit that the
issue could have been readily solved considering the basic antenna theory.

We know that two stacked dipoles yield a gain of up to 3 dB with respect to a
single dipole. In other words, two dipoles each fed with power P/2 produce, in
some regions of the space, a higher field than that produced by a single dipole
fed with power P, although the total transmit power has not changed.

The extra power measured at the receiver is obviously "created" at the expense
of power taken away from other regions of the space (according to the transmit
antenna pattern).

Too fundamental to deserve further discussions!

73

Tony I0JX

Antonio Vernucci wrote:
The extra power measured at the receiver is obviously "created" at the
expense of power taken away from other regions of the space (according
to the transmit antenna pattern).

Too fundamental to deserve further discussions!

Almost everyone knows what occurs in free space -
constructive interference in the direction of
greater gain and destructive interference in the
direction of lesser gain. But my posting was not
about free space. I thank you for your input so
far but please now extend those EM wave concepts
to transmission lines.

Everyone doesn't agree that constructive and
destructive interference also happens at a Z0-
match point in a transmission line with reflections.
That is the topic that needs "further discussions".

Just as constructive interference functions to
increase antenna gain in one direction while
destructive interference functions to decrease
antenna gain in another direction, in a transmission
line at a Z0-match point, constructive interference
functions to increase the energy flow toward the
load while destructive interference functions to
decrease the energy flow toward the source.

Antonio, please don't bow out now. You are apparently
one of the few posters who fully understands interference.
--
73, Cecil http://www.w5dxp.com