I wrote:
"With a perfect ground in this situation (MW vertical tower) the base
current in the vertical mast is twice what it would be for the same
power applied to an equivalent wire dipole (less a ground system) in
free space -- resulting in 3 dB system gain."
to which Richard Harrison responded.
____________

Excuse my inaccurate statement about that current, and thanks for catching
it.

The input resistance of a 1/4-wave monopole working against a perfect ground
plane is 36.5 ohms,* or half that of a 1/2-wave dipole in free space. A
given input power then results in 1.414X more current in the monopole than
the dipole. Hence the monopole radiates 1.414X the field of the dipole.
And, as shown in Kraus 3rd edition, Table 6-2, a 1/4-wave monopole against a
perfect ground has 3 dB more gain than a 1/2-wave dipole in free space.

Increasing the field from the 1/2-wave, free space dipole by 1.414X in the
above example would require doubling its input power (3dB), which would
result in the same 1.414X increase in its current -- to the same value as
seen in the monopole with 1/2 that power.

* per Kraus 3rd Edition, p 567

RF

This is unfortunately an example of arriving at the right result by
using the wrong (actually, incomplete) method.

The field from a conductor is proportional not only to the current
flowing in it, but also the length of the conductor. So if you put the
same current into a dipole and monopole, and assuming they have the same
current distribution, the field from the dipole will be twice the field
from the monopole.

So let's start again. The input R of a monopole over an infinite perfect
ground is 1/2 the resistance of a free space dipole, so for a given
power input the monopole current is 1.414 times the dipole current, as
you said. Ok so far. But because the dipole is twice as long and with
the same current distribution (and oriented in such a way that the
fields from the two halves add in phase), the field from the dipole is
2/1.414 = 1.414 times the field from the monopole. However, each ray
from the monopole is reflected from ground, resulting in two rays adding
in phase at a distant point. This doubles the field from the monopole,
so it's now 2/1.414 = 1.414 times the field from the free space dipole.
This is the same result, but with the two additional important factors
of radiator length and ground reflection included.

A good check of the final result is to note that, neglecting loss, the
average field intensity from *any* antenna in free space has to be 3 dB
less than the average field intensity from *any* antenna over an
infinite ground plane, if the same power is applied to each. The reason
is simply that the supplied power is spread over half the volume when
the ground plane is present.

Roy Lewallen, W7EL

Richard Fry wrote:
. . .
The input resistance of a 1/4-wave monopole working against a perfect
ground plane is 36.5 ohms,* or half that of a 1/2-wave dipole in free
space. A given input power then results in 1.414X more current in the
monopole than the dipole. Hence the monopole radiates 1.414X the field
of the dipole. And, as shown in Kraus 3rd edition, Table 6-2, a 1/4-wave
monopole against a perfect ground has 3 dB more gain than a 1/2-wave
dipole in free space.

Increasing the field from the 1/2-wave, free space dipole by 1.414X in
the above example would require doubling its input power (3dB), which
would result in the same 1.414X increase in its current -- to the same
value as seen in the monopole with 1/2 that power.

* per Kraus 3rd Edition, p 567

RF