"gareth" wrote in message
...
Actually, here is a better example, because it represents the situation
found in many shacks.
Consider 100W at 3.6MHz propagating along some 50 ohm
coax, which terminates suddenly but with 1/4 inch of the central
conductor protruding, and thereby forming a short antenna.
The short antenna, only 1/4 inch long is immediately terminated
by a 50 ohm resistance.
1. How much of the power from the coax is fed into that short antenna
despite
the claimed (by others) impedance mismatch?
2. How much of that power is radiated by that short antenna?
3. If all the power that is fed to the short antenna is radiated, does
the
50 ohm resistor dissipate any of it?
4. How much of the power is dissipated in the 50 ohm resistor?
5. How much of the power is reflected back down the coax because
of the impedance mismatch of that (very) short antenna?
6. Of those who claim that a short antenna will radiate all the power fed
to it,
how many will realise that for any power to be dissipated in the resistor,
it must
have been successfully fed to that short antenna in the first place?
7. Of those who suggest that impednace matching is a serious
consideration, how
many will realise that at 3.6MHz, that the 1/4" short antenna is the
standard practice to
connect the end of the coax to the dummy load with a bit of wire?

Setting aside dielectric and slot antennae, an antenna is a conductor into
which
power has been passed.

Once the power has entered that antenna, how it got there by feed, and how
the
antenna is terminated are irrelevant. In the example above, the short piece
of
wire coupling the coax to a dummy load is a short antenna, and yet it
clearly is
not radiating all the power being fed to it, as most of it will be
dissipated in the
dummy load.

Nevertheless, it is a valid example of power being fed to a short antenna
which is
not all being radiated, because a short antenna does not radiate
efficiently, which is
where we came in.

"gareth" wrote in message ...

Consider 100W at 3.6MHz propagating along some 50 ohm
coax, which terminates suddenly but with 1/4 inch of the central
conductor protruding.

Now there's no difficulty in feeding all that power into that
1/4 inch because it is so short compared to a wavelength
that there is a uniformity of voltage and current along it,
and it will be essentially the same as that existing in the last
gnat's cock of the coax.

Attach a hi-impedance scope probe to the end of that
1/4 inch and all the power being delivered through the
coax will be detectable right at the tip of that 1/4 inch.

Now, will that 1/4 inch antenna radiate all the power that
is being successfully fed to it at 3.6MHz, or will the
configuration behave merely as an open-circuit with all
the power being reflected back down the coax?

A number of contributors to this NG claim that the 1/4 inch
stub antenna will radiate the full 100W at 3.6MHz,
snip

In this example, the transmitter delivers 100w to the coax, but only a small
part of that is delivered to the 1/4 inch "antenna".

The 1/4 inch will radiate the power delivered to it just as well as a to a
full sized antenna.

Matching and radiation are two different subjects.

gareth wrote:

"gareth" wrote in message
...

Actually, here is a better example, because it represents the situation
found in many shacks.

Consider 100W at 3.6MHz propagating along some 50 ohm
coax, which terminates suddenly but with 1/4 inch of the central
conductor protruding, and thereby forming a short antenna.

The short antenna, only 1/4 inch long is immediately terminated
by a 50 ohm resistance.

1. How much of the power from the coax is fed into that short antenna
despite
the claimed (by others) impedance mismatch?

2. How much of that power is radiated by that short antenna?

3. If all the power that is fed to the short antenna is radiated, does the
50 ohm resistor dissipate any of it?

4. How much of the power is dissipated in the 50 ohm resistor?

5. How much of the power is reflected back down the coax because
of the impedance mismatch of that (very) short antenna?

6. Of those who claim that a short antenna will radiate all the power fed to
it,
how many will realise that for any power to be dissipated in the resistor,
it must
have been successfully fed to that short antenna in the first place?

Nope.

Consider a 100V source, a 1 Ohm resistor and a 100 Ohm resistor in series.

The power dissipated by the 1 Ohm resistor is 0.99 W and the power
dissipated in the 100 Ohm resistor is 99 W.

Phrases like "power fed to it" are meaningless word salad.

You can apply voltage and you can apply current, but you can NOT feed power.


7. Of those who suggest that impednace matching is a serious consideration,
how
many will realise that at 3.6MHz, that the 1/4" short antenna is the
standard practice to
connect the end of the coax to the dummy load with a bit of wire?

Non sequitur.


--
Jim Pennino

gareth wrote:
"gareth" wrote in message
...
Actually, here is a better example, because it represents the situation
found in many shacks.
Consider 100W at 3.6MHz propagating along some 50 ohm
coax, which terminates suddenly but with 1/4 inch of the central
conductor protruding, and thereby forming a short antenna.
The short antenna, only 1/4 inch long is immediately terminated
by a 50 ohm resistance.
1. How much of the power from the coax is fed into that short antenna
despite
the claimed (by others) impedance mismatch?
2. How much of that power is radiated by that short antenna?
3. If all the power that is fed to the short antenna is radiated, does
the
50 ohm resistor dissipate any of it?
4. How much of the power is dissipated in the 50 ohm resistor?
5. How much of the power is reflected back down the coax because
of the impedance mismatch of that (very) short antenna?
6. Of those who claim that a short antenna will radiate all the power fed
to it,
how many will realise that for any power to be dissipated in the resistor,
it must
have been successfully fed to that short antenna in the first place?
7. Of those who suggest that impednace matching is a serious
consideration, how
many will realise that at 3.6MHz, that the 1/4" short antenna is the
standard practice to
connect the end of the coax to the dummy load with a bit of wire?

Setting aside dielectric and slot antennae, an antenna is a conductor into
which
power has been passed.

Once the power has entered that antenna, how it got there by feed, and how
the

Power does not enter an antenna, voltage and/or current enter an antenna.

antenna is terminated are irrelevant. In the example above, the short piece
of
wire coupling the coax to a dummy load is a short antenna, and yet it
clearly is
not radiating all the power being fed to it, as most of it will be
dissipated in the
dummy load.

As Ohms law says it should because of the impedance of the antenna compared
to the impedance of the resistive load.

Nevertheless, it is a valid example of power being fed to a short antenna

There is no such things as "power being fed to" anything; you can only
apply voltage or current.

The only place were power is fed is in Star Trek.

which is
not all being radiated, because a short antenna does not radiate
efficiently, which is

You have been shown repeatedly that this is false.


--
Jim Pennino

Jeff wrote:

Actually, here is a better example, because it represents the situation
found in many shacks.

Consider 100W at 3.6MHz propagating along some 50 ohm
coax, which terminates suddenly but with 1/4 inch of the central
conductor protruding, and thereby forming a short antenna.

The short antenna, only 1/4 inch long is immediately terminated
by a 50 ohm resistance.

1. How much of the power from the coax is fed into that short antenna
despite
the claimed (by others) impedance mismatch?

2. How much of that power is radiated by that short antenna?

3. If all the power that is fed to the short antenna is radiated, does the
50 ohm resistor dissipate any of it?

4. How much of the power is dissipated in the 50 ohm resistor?

5. How much of the power is reflected back down the coax because
of the impedance mismatch of that (very) short antenna?


It depends on how you connect the 50ohm load. The ground side of the 50
ohm load must go somewhere, so logic would dictate that it was connected
to the outer of the coax. If that case the 1/4 inch is so small compared
to a wavelength that it will not present much of a mismatch to the coax
at 3.6MHz, the coax will see 50ohms in series with a very small
inductance and even smaller stray capacitance, virtually all of the
power will be dissipated in the load, a small ammopunt will be
reflected. ie you have just added a small lumped inductor in series with
the 50ohms.

To be pendatic, the short wire has a small resistance and a small impedance,
both of which are in milliohms.

Since P=I^2R and the R is very, very small, the P dissipated both by
radiation and heat are very small.


This example has nothing to do with how short whips radiate. You keep
harping back to feed lines which are not what is under discussion. We
all agree that practically matching short whips is where the problem
lies. However, there is unequivocal proof that IF power is got into a
short whip by some means or other it is ALL radiated or lost as heat in
the resistance of the element.

Read Kraus, Jasik, or do some NEC simulations yourself and you will see
and hopefully stop this ridiculous quest for an answer that is just not
there.

Jeff



--
Jim Pennino

Wayne wrote:


"gareth" wrote in message ...

Consider 100W at 3.6MHz propagating along some 50 ohm
coax, which terminates suddenly but with 1/4 inch of the central
conductor protruding.

Now there's no difficulty in feeding all that power into that
1/4 inch because it is so short compared to a wavelength
that there is a uniformity of voltage and current along it,
and it will be essentially the same as that existing in the last
gnat's cock of the coax.

Attach a hi-impedance scope probe to the end of that
1/4 inch and all the power being delivered through the
coax will be detectable right at the tip of that 1/4 inch.

Now, will that 1/4 inch antenna radiate all the power that
is being successfully fed to it at 3.6MHz, or will the
configuration behave merely as an open-circuit with all
the power being reflected back down the coax?

A number of contributors to this NG claim that the 1/4 inch
stub antenna will radiate the full 100W at 3.6MHz,
snip

In this example, the transmitter delivers 100w to the coax, but only a small
part of that is delivered to the 1/4 inch "antenna".

The 1/4 inch will radiate the power delivered to it just as well as a to a
full sized antenna.

Matching and radiation are two different subjects.

Correct but it is much easier to see if one looks at it in terms of
voltage and current and calculates the power at any given point.

The equivelant circuit to a 100 W, 50 Ohms transmitter is a 141.4 V
voltage source with a 50 Ohm resistor.

The impedance of this antenna is a very small value, assume 0.001 Ohms
as a reasonable value for analysis.

There is 141.4 V applied to 50.001 Ohms, so the current is 141.4/50.001.

The power in the antenna is I^2R, which is (141.4/50.001)^2*0.001,
which my calculator says is 0.008 W.

Where is the other 99.992 W?

It is dissipated in the internal resistance of the transmitter and
given real world coax, some of it is dissipated as heat in the coax
because of the huge VSWR.


--
Jim Pennino