Jim Kelley wrote:
Were it not for waves standing on the coax, there would
have been no damage to the coax, ...

Were it not for traveling waves, there would be no
standing waves and no damage to the coax.

Kinda tough to have a standing wave in the absence of traveling waves
now, isn't it.

Exactly, but that supports my side of the argument.

So there's energy in the standing waves, but not power.

At the risk of you developing apoplexy, there is reactive
power in the standing waves, as defined in The IEEE Dictionary.

Probably best not to assume it's a 'different kind of wave' in the first
place.

Sorry Jim, but [Io*cos(kx)*cos(wt)] and [Io*cos(kx+wt)] *ARE*
different. If you don't know enough math to realize that by
looking at the equations, please go alleviate your ignorance.

Yes I know that's basically what you've been trying to say "all along",
but you were saying it so poorly ...

Sorry, my native tongue is Texan. :-)
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:

Sorry Jim, but [Io*cos(kx)*cos(wt)] and [Io*cos(kx+wt)] *ARE*
different.

Evidently, you can't recognize a trig identity when you see one.

If you don't know enough math to realize that by
looking at the equations, please go alleviate your ignorance.

Sorry, my native tongue is Texan. :-)

Your tongue may be Texan, but the rest of you is pure arsehole. :-)

ac6xg

Jim Kelley wrote:
Cecil Moore wrote:
Sorry Jim, but [Io*cos(kx)*cos(wt)] and [Io*cos(kx+wt)] *ARE*
different.

Evidently, you can't recognize a trig identity when you see one.

Good grief, Jim, please solve the following "identity".

cos(kx)*cos(wt) = cos(kx+wt)

Find all (x,t) for which that equation is true.
I'm expecting an "attitude adjustment" from you.
Everyone is invited to solve the above equation
for (x,t). Hint: I solved it a long time ago.
--
73, Cecil http://www.w5dxp.com

"Jim Kelley" wrote in message
...
Cecil Moore wrote:

Sorry Jim, but [Io*cos(kx)*cos(wt)] and [Io*cos(kx+wt)] *ARE*
different.

Evidently, you can't recognize a trig identity when you see one.

in my book
cos(a+b)=cos(a)cos(b)-sin(a)sin(b)
so cos(kx+wt) would expand to:
cos(kx)cos(wt)-sin(kx)sin(wt)
me thinks you are missing a few terms in your 'identity'.

Dave wrote:
"Jim Kelley" wrote in message
...

Cecil Moore wrote:


Sorry Jim, but [Io*cos(kx)*cos(wt)] and [Io*cos(kx+wt)] *ARE*
different.

Evidently, you can't recognize a trig identity when you see one.


in my book
cos(a+b)=cos(a)cos(b)-sin(a)sin(b)
so cos(kx+wt) would expand to:
cos(kx)cos(wt)-sin(kx)sin(wt)
me thinks you are missing a few terms in your 'identity'.

I'm sure your book is correct, Dave. Mine probably is too. (It has
CRC Standard Mathematical Tables printed on the cover.) Cecil through
a curve ball. The correct function for a standing wave can be written
as the algebraic sum of two sine functions, or as the product of a
sine and a cosine function e.g. y = 2*Ymax*sin(kx)cos(wt).

73, ac6xg

Jim Kelley wrote:
I'm sure your book is correct, Dave. Mine probably is too. (It has CRC
Standard Mathematical Tables printed on the cover.) Cecil through a
curve ball. The correct function for a standing wave can be written as
the algebraic sum of two sine functions, or as the product of a sine and
a cosine function e.g. y = 2*Ymax*sin(kx)cos(wt).

I'm surprised that a physics professor doesn't recognize
the difference between the conventions of RF engineering
and optical physics (which are essentially meaningless).

Is discrediting me really worth denying the laws of physics?
--
73, Cecil http://www.w5dxp.com

Jim Kelley wrote:
Cecil Moore wrote:

Sorry Jim, but [Io*cos(kx)*cos(wt)] and [Io*cos(kx+wt)] *ARE*
different.

Evidently, you can't recognize a trig identity when you see one.

One really should take a look at the math before waving one's
hands and opening one's mouth in ignorance.

Please enlighten us as to exactly what trig "identity" will make
the following terms equal.

E1*e^j(wt-kx) ?=? E2*e^j(wt-kx) + E2*e^j(wt+kx)

Seems to me the only condition for which they are equal is when
E2=0, i.e. when reflections (and therefore standing waves) don't
exist.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Jim Kelley wrote:

Cecil Moore wrote:

Sorry Jim, but [Io*cos(kx)*cos(wt)] and [Io*cos(kx+wt)] *ARE*
different.


Evidently, you can't recognize a trig identity when you see one.


One really should take a look at the math before waving one's
hands and opening one's mouth in ignorance.

Please enlighten us as to exactly what trig "identity" will make
the following terms equal.

E1*e^j(wt-kx) ?=? E2*e^j(wt-kx) + E2*e^j(wt+kx)

Seems to me the only condition for which they are equal is when
E2=0, i.e. when reflections (and therefore standing waves) don't
exist.

The purpose of pointing out the trigonometric relationship between the
sum of sines and the product of sine and cosine was to illustrate
that, contrary to your assertion, there isn't a difference in the
waves. The traveling waves can either be written mathematically as
two separate traveling waves, or as one standing wave. It makes no
difference; the waves are the same in either case irrespective of how
you choose to describe them mathematically. Do you grasp the meaning
here, or not?

Thanks,

ac6xg

Jim Kelley wrote:
The purpose of pointing out the trigonometric relationship between the
sum of sines and the product of sine and cosine was to illustrate that,
contrary to your assertion, there isn't a difference in the waves.

Jim, please go ask the head of your math department if
there's any difference in those equations. That you don't
see any difference is just extreme ignorance on your part.

Set cos(kx-wt) = cos(kx)*cos(wt) and wrestle with the
trig identities until you alleviate your ignorance.

You are extremely wrong and ignorant of mathematics.
You will realize that fact when you are unable to
prove your assertions even to yourself.

Hint: cos(kx+wt) = cos(kx)*cos(wt) - sin(kx)*sin(wt)
You are obviously missing half of the terms when you
say there "isn't a difference in the waves". (FYI,
anyone who knows anything about mathematics is laughing
at you.)
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:

That you don't
see any difference is just extreme ignorance on your part.

Set cos(kx-wt) = cos(kx)*cos(wt) and wrestle with the
trig identities until you alleviate your ignorance.

Cecil,

There are an infinite number of different functions one could write to
describe an infinite number of different possible wave shapes. And
none of them would necessarily be mathematically equivalent to
another. But when you write the equation for the superposition of
traveling waves and claim that resultant standing wave is a different
kind of electromagnetic wave, that is a misguided point of view. The
equation for a standing wave is simply a different way of writing the
sum of two traveling waves. Being that there is only one kind to
chose from, there cannot be a different kind of electromagnetic wave.
Is it impossible for you to acknowledge this simple point in a
gentlemanly fashion?

You are extremely wrong and ignorant of mathematics.

Any 'extreme wrongness' notwithstanding, what I don't know about
mathematics could fill a book. On the other hand, you almost
inevitably end up lying and turning technical discussions into
personal attacks. Ask anyone.

ac6xg