tom wrote:
On 4/30/2013 3:17 AM, Szczepan Bialek wrote:
"tom" napisal w wiadomosci
. ..
On 4/29/2013 10:20 PM, tom wrote:
On 4/29/2013 3:50 AM, Szczepan Bialek wrote:
"Wayne" napisal w wiadomosci
Assuming the monopole is 1/4 wave long, and has a 1/4 wave long radial.
The radial is in line with the monopole. Everything is ungrounded.

The radial connected with the shield of the coax is the ground.

So what would happen if I connected my transmitter, which has a 50 ohm
{ostensibly) output with the hot lead of the coax connected to the
"radial" instead of the "monopole" and similarly with the braid?

Think long about this.


Just to clearly understand where you are on how this really works.

I am not sure if I understand you.

If you have the mechanically symmetric dipole than one leg with the coax
braid works as the radiator and the second as the ground.
If you change the terminals than your radial will be the radiator.

Do not you have any possibility to check which leg is active?
S*



How does the "leg" know if it is connected to the braid versus the
center conductor?

tom
K0TAR

A better question is how does this idiot to come in out of the rain
or does he just stand there, looking at the sky with with his mouth open,
until someone notices the gurgling sounds and takes him inside?





--
Jim Pennino