In message , Szczepan Bialek
writes

"Rob" napisa³ w wiadomo¶ci
...
Szczepan Bialek wrote:

napisa3 w wiadomo?ci
...
Szczepan Bialek wrote:

"Ian Jackson" napisal w
wiadomosci
...

In the past, many amateurs did connect coax directly to a dipole.

And what they do if they have the monopole?

Connect it with coaxial transmission line, idiot.

If the monopole has only one radial it is exactly as the your dipole.
Right?

No. A monopole needs many radials, not one. if it has only one radial
the voltage at the end of the radial is the same as on the monopole,
and the whole thing becomes a dipole.

"balun definition
electronics
A transformer connected between a balanced source or load and an unbalanced
source or load.

Correct.

A balanced line has two conductors, with equal currents in
opposite directions.

Correct.

The unbalanced line has just one conductor; the current
in it returns via a common ground or earth path. "

Incorrect.

Unbalanced line (ie coax) has two conductors - the inner conductor and
the screen. At RF, the go and return currents flow on the outside if the
inner conductor, and on the inside of the screen. The outside of the
screen should be 'RF dead'. However, when the unbalanced coax is
connected directly to a balanced antenna, a lot of the screen current
flows on the outside of the screen, so the coax becomes part of the
antenna, and radiates.

The braid of the coax and the radial are connected. They work as the ground
(or counterpoise).

A counterpoise (or a system of radials) can provide an artificial
ground. However, an artificial ground is only required for 'one-sided'
monopole types of antennas. A counterpoise is not normally intended to
be a radiator.

The 'counterpoise' side of a dipole - if it is connected directly to the
coax screen - is not RF-dead. It is usually intended to radiate
more-or-less as well as the 'RF-live' side. Unfortunately, some of the
current in the 'counterpoise' side of the dipole flows down the outside
of the coax screen, and this causes the coax to radiate.

Is the voltage the same as in the live conductor (at the feed points)?

Which voltages do you mean?
--
Ian