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

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

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

*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?
--
73, Cecil http://www.w5dxp.com

"Cecil Moore" wrote in message
et...
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

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

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

*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?
--
73, Cecil http://www.w5dxp.com

Cecil

Congratulations, you appear to have solved the world's energy problems for
ever! Electricity will now be too cheap to meter, just like they promised
when nuclear power plants were first built.

Mike G0ULI

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

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

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

*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?

The peak intensity of a standing wave will always be greater than the
simple sum of the two waves. Itot = 4*I*cos^2(deg/2). But I think
nature will somehow conspire against you if you try to make use of
more than the 100 watts input. ;-)

ac6xg

On Nov 16, 12:34 pm, 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

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

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

*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?
--
73, Cecil http://www.w5dxp.com

Nice "when are you going to stop beating your mother" sort of
question. And what was your reply?

"Cecil Moore" wrote in message
et...
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:

Can we take this to mean that this is not the steady state condition?

Tam/WB2TT

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

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

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

*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?
--
73, Cecil http://www.w5dxp.com

Mike Kaliski wrote:

"Cecil Moore" wrote in message
*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?

Congratulations, you appear to have solved the world's energy problems
for ever! Electricity will now be too cheap to meter, just like they
promised when nuclear power plants were first built.

Sorry, unlike others on this newsgroup, I don't support
a violation of the conservation of energy principle. To
answer my own question above, the extra 71 joules/sec
of constructive interference comes from 71 joules/sec
of destructive interference occurring somewhere else.
--
73, Cecil http://www.w5dxp.com

Jim Kelley wrote:
The peak intensity of a standing wave will always be greater than the
simple sum of the two waves.

No reference to peaks here, Jim. Everything is average values.
The average value of the energy in the standing waves *always*
equals the average values of the energy components of the
forward and reflected waves added together. If there are X joules
in the standing waves, there will be X joules in the sum of the
forward and reflected waves.

But I think
nature will somehow conspire against you if you try to make use of more
than the 100 watts input. ;-)

It was a rhetorical question for people who say, "Just do
a vector analysis and the energy will take care of itself."
Understanding exactly how the energy takes care of itself
is the point of this thread. Hint: 171 joules/sec from two
50 joules/sec waves requires an additional source of energy.
In the case of this example, the 71 joules/sec of constructive
interference requires 71 joules/sec of destructive interference
energy occurring somewhere else.
--
73, Cecil http://www.w5dxp.com

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

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

50 joules/s are carried by Wave#1 if alone. The same applies to Wave#2

But if both waves are sumultaneouly present, the power carried by each wave when
alone is no longer a meaningful number. As a matter of fact when superposing two
coherent waves (same frequency, fixed phase relationship), one MUST first sum
voltages (or currents) and then calculate power.

Summing wave powers could only be done in case of incoherent waves.

In conclusion, the answer to your question is that the apparent extra 71
joules/s come front the fact that 100 joules/s taken as reference is a number
having no physical meaning.

73

Tony I0JX

K7ITM wrote:
Nice "when are you going to stop beating your mother" sort of
question. And what was your reply?

It's a rhetorical question, Tom. What is your reply?
When someone (besides Eugene Hecht) explains it to
my satisfaction I will stop beating that dead horse.
--
73, Cecil http://www.w5dxp.com

Tam/WB2TT wrote:
"Cecil Moore" wrote:
Given two coherent EM waves superposed in a 50 ohm environment
at considerable distance from any source:

Can we take this to mean that this is not the steady state condition?

Sorry if I somehow gave you that idea. It is a steady-state
problem.
--
73, Cecil http://www.w5dxp.com