wrote:
Radiation does not cause current taper. Dissipation does not either.

Radiation and dissipation are considered to be losses in a
transmission line covered by the attenuation factor. All that
is needed to prove your above assertions to be false is to
quote a transmission line equation. It can even be the more
simple flat form where the SWR is 1:1. Here it is in ASCII:

I = Im*e^(ax)*e^j(wt-bx)

Note this is the equation for *CURRENT* where 'a' is the attenuation
factor. The attenuation factor includes radiation and dissipation.

Your statements indicate a high level of ignorance. Assuming a flat
transmission line with an SWR of 1:1, if the loss in the transmission
line is 3 dB, we can put 200 watts into 50 ohm coax at the source end
and get 100 watts out at the 50 ohm load end. The current out of the
source is SQRT(200w/50) = 2 amps. The current through the 50 ohm load
is SQRT(200w/50) = 1.414 amps. The current has dropped from the source
to the load by exactly the same percentage that the voltage has dropped.

What you seem to be missing is that the H-field is attenuated by the
same amount as the E-field while the ratio of E-field to H-field
remains constant and equal to Z0. Current is proportional to the
H-field and voltage is proportional to the E-field.
--
73, Cecil http://www.qsl.net/w5dxp

Cecil Moore wrote:

wrote:

Radiation does not cause current taper. Dissipation does not either.


Radiation and dissipation are considered to be losses in a
transmission line covered by the attenuation factor. All that
is needed to prove your above assertions to be false is to
quote a transmission line equation. It can even be the more
simple flat form where the SWR is 1:1. Here it is in ASCII:

I = Im*e^(ax)*e^j(wt-bx)

Note this is the equation for *CURRENT* where 'a' is the attenuation
factor. The attenuation factor includes radiation and dissipation.

Your statements indicate a high level of ignorance. Assuming a flat
transmission line with an SWR of 1:1, if the loss in the transmission
line is 3 dB, we can put 200 watts into 50 ohm coax at the source end
and get 100 watts out at the 50 ohm load end. The current out of the
source is SQRT(200w/50) = 2 amps. The current through the 50 ohm load
is SQRT(200w/50) = 1.414 amps. The current has dropped from the source
^^^
Obviously, should be 100w. Sorry for the typo.

to the load by exactly the same percentage that the voltage has dropped.

What you seem to be missing is that the H-field is attenuated by the
same amount as the E-field while the ratio of E-field to H-field
remains constant and equal to Z0. Current is proportional to the
H-field and voltage is proportional to the E-field.
--
73, Cecil http://www.qsl.net/w5dxp

Cecil,

Wow! I think you may have set a new world record for the most irrelevant
concepts per word dragged into an RRAA posting.

We got transmission lines, attenuation factors, H-fields, E-fields, and
even SWR. Not to mention watts, dB, and Zo.

It is truly unfortunate that none of this is connected to the subject at
hand, displacement current, but it makes for a colorful message.

73,
Gene
W4SZ

Cecil Moore wrote:
wrote:

Radiation does not cause current taper. Dissipation does not either.


Radiation and dissipation are considered to be losses in a
transmission line covered by the attenuation factor. All that
is needed to prove your above assertions to be false is to
quote a transmission line equation. It can even be the more
simple flat form where the SWR is 1:1. Here it is in ASCII:

I = Im*e^(ax)*e^j(wt-bx)

Note this is the equation for *CURRENT* where 'a' is the attenuation
factor. The attenuation factor includes radiation and dissipation.

Your statements indicate a high level of ignorance. Assuming a flat
transmission line with an SWR of 1:1, if the loss in the transmission
line is 3 dB, we can put 200 watts into 50 ohm coax at the source end
and get 100 watts out at the 50 ohm load end. The current out of the
source is SQRT(200w/50) = 2 amps. The current through the 50 ohm load
is SQRT(200w/50) = 1.414 amps. The current has dropped from the source
to the load by exactly the same percentage that the voltage has dropped.

What you seem to be missing is that the H-field is attenuated by the
same amount as the E-field while the ratio of E-field to H-field
remains constant and equal to Z0. Current is proportional to the
H-field and voltage is proportional to the E-field.

Gene Fuller wrote:
It is truly unfortunate that none of this is connected to the subject at
hand, displacement current, but it makes for a colorful message.

Please don't tell us that you don't understand how the attenuation
factor in a transmission line current equation causes the current
to drop along the line equaling the percentage drop in the voltage.

One can write a similar equation for a standing wave dipole.
--
73, Cecil http://www.qsl.net/w5dxp