"Cecil Moore" wrote in message
news
Antonio Vernucci wrote:
Each wave produces 50 joules/s when alone. When the two waves are
superimposed, each wave produces not only its 50 joules/s but also 35.5
more joules that it "robs" from other regions of the space.

My point exactly! The same thing is true when it happens
in a transmission line at a Z0-match point. The region of
constructive interference toward the load (forward energy
wave) "robs" energy from the region of destructive
interference toward the source (reflected energy waves).
That's how antenna tuners work.
--
73, Cecil http://www.w5dxp.com

You can come up with a lot simpler example that at first might look like a
paradox. Consider two DC current sources of 1 amp each. Each current source
will deliver 50 W to a 50 Ohm resistor. Now connect the two current sources
in parallel, and the resultant 2 amps will deliver 200W to the same 50 Ohm
resistor. There is nothing wrong here.

Tam

Cecil Moore wrote:
The following is from an email to which I replied today.

Given two coherent EM waves superposed in a 50 ohm environment
at considerable distance from any source:

Wave#3 = Wave#1 superposed with Wave#2

Wave#1: V = 50v at 0 deg, I = 1.0a at 0 deg, P = 50 joules/sec

yes, lets say source S1 supplies a voltage V1 into a load L1, where L1
is a pure 50 Ohm resistance.


Wave#2: V = 50v at 45 deg, I = 1.0a at 45 deg, P = 50 joules/sec

Source S2 supplies a voltage V2 puts into a load L2, where L2 is a pure
50 Ohm resistance.

These two waves superpose to V = 92.38v and I = 1.85a
Note: P = 171 joules/sec

You have changed the circuit.

Source S1 is no longer connected to a load L1 consisting of a 50 Ohm
load. It is connected to a load L3, consisting of a pure resistance in
series with a voltage source.

Since you have changed the circuit source 1 is connected to, you should
not be surprised it supplies a different power.

Move the phase difference to 180 degrees, and source S1 would supply no
power at all.


*During each second*, Wave#1 supplies 50 joules of energy and
Wave#2 supplies 50 joules of energy for a total of 100 joules of
energy being supplied *every second* to the superposition process.
Yet the results of that superposition process yields 171 joules
of energy *during each second*, 71 joules more than is being
supplied to the process. Where are the extra 71 joules per second
coming from?

35.5 J/s (Watts) comes from the source S1 and 35.5 J/s (Watts) comes
from the source S2. It's different to the first case, as they are
connected to a different circuit.

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
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.

Dave wrote:
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.


Sorry, 0 or 2A. Then the power is 0 or 200W as I said.

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

Dave wrote:
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.

Piece of cake. The sources are beams of light from Alpha Centauri.
Now tell us exactly how Alpha Centauri adjusts its power output
from 11 light-years away.
--
73, Cecil http://www.w5dxp.com