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

On Fri, 04 Jan 2008 14:31:33 -0600, Cecil Moore
wrote:

If you are going to excite my example remotely, you
need a 411 ohm load resistor on each end of the wire.

:-O

There IS a Santa Claus! Thank you, Thank you, Thank YOU!

I did that too!!!!

SWR = 29:1

Dear Readers,

The snap of the whip sings holiday carols when you hold back your
trump cards. A late present, and a most wonderful one. ;-)

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

Richard Clark wrote:
Cecil Moore wrote:
If you are going to excite my example remotely, you
need a 411 ohm load resistor on each end of the wire.

SWR = 29:1

You are mistaken or bearing false witness - neither
of which is a plus for you.
--
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'.

Richard Clark wrote:
SWR = 29:1

Richard, to prove that you are not 100% insane,
would yo please attempt to prove that the SWR
in the transmit antenna is in any way related
to the SWR in the receive antenna?
--
73, Cecil http://www.w5dxp.com

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

On Sun, 6 Jan 2008 17:37:37 -0500, "AI4QJ" wrote:


"Richard Clark" wrote in message
.. .
On Fri, 04 Jan 2008 20:03:04 GMT, Cecil Moore
wrote:

The SWR on the wire is load-
based and is equal to
8:1 as evidenced by CURRENT on the wire. :-)

You cannot make a SWR measurement on a receive antenna any other way.

You are making a "Voltage" standing wave measurement based ion the CURRENT
of the wire?

Hi Dan,

I don't know about ions; but, sure, voltage standing wave measurement
is as accurate as current standing wave measurement. Or perhaps you
need to make your question a little more terse (don't rely on the
continuity of discussion involving Cecil's quotes to fully inform any
argument).

In other words, filter out the static and frame your own question.

73's
Richard Clark, KB7QHC

"AI4QJ" wrote in message
...

"Richard Clark" wrote in message
...
On Fri, 04 Jan 2008 20:03:04 GMT, Cecil Moore
wrote:

The SWR on the wire is load-
based and is equal to
8:1 as evidenced by CURRENT on the wire. :-)

You cannot make a SWR measurement on a receive antenna any other way.

You are making a "Voltage" standing wave measurement based ion the CURRENT
of the wire?

AI4QJ

sure, VSWR is just as valid as Current-SWR, they are absolutely identical...
the only difference is that VSWR is the amateur 'standard' because it was
frequently easier to measure the voltage than the current... i.e. on open
wire feeders all you had to do was run a neon bulb or other cheap voltmeter
along the line to find the min and max voltage points... for a current
measurement you would have to couple a loop around the line or insert
shunts, not nearly as simple as measuring a voltage... so vswr was the
'standard'... though just plain swr, refering to either one, is just as
proper.