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, but
I fear that they are sadly mistaken and that their associated
infantile outbursts are because they are in denial either
about their error, or cannot face up to a challenge to
their seemingly-religious faith as to what is happening.

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

A piece of coax is not an antenna.

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.

Wrong and shows an utter lack of understanding how antennas work.

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.

But not the current in that 1/4 inch piece, which is the important part.

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?

The amount of power the will be successfully fed to it at 3.6MHz will
be miniscule.

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

If and only if the impedance of the 1/4 inch stub is matched to 50 Ohms
without losses in the matching device and the resistance of the 1/4 inch
stub is much less than the resistive input impedance of the 1/4 inch
stub.

This is something you just can not seem to understand no matter how
many times it is repeated or shown to be true.

but
I fear that they are sadly mistaken and that their associated
infantile outbursts are because they are in denial either
about their error, or cannot face up to a challenge to
their seemingly-religious faith as to what is happening.

I fear you are incapable of understanding how any antenna works.




--
Jim Pennino

"Brian Reay" wrote in message
...

He doesn't want to understand, he wants to stir up a long thread he can
pepper with abuse

Untrue.

and
claim he 'won',

Untrue. Once again you project your own failings. Also untrue, because I
am always willing to accept that I may be wrong, unlike yourself who
regularly changes the subject in order to avoid being challenged; a
strategy for which you are notorious.

while others who disagreed with him didn't understand the
subject.

Untrue, although they might have misconstrued and shot off at
a tangent, as you are wont to do.

As strange as it sounds, this seems to give him some kind of 'buzz',
perhaps his self esteem is so low that he
is driven to such measures.

Untrue.

Perhaps he is just a nasty individual.

Physician, heal thyself.

Once again, Brian, you have jumped into a technical thread only to
originate abusive remarks but without making any technical contribution.

Why do you behave like that?

gareth wrote:
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, but
I fear that they are sadly mistaken and that their associated
infantile outbursts are because they are in denial either
about their error, or cannot face up to a challenge to
their seemingly-religious faith as to what is happening.

The resistive impedance of your 1/4 inch stub will be a small fraction
of an Ohm.

Let us assume 0.001 Ohms though it would likely be much smaller than
that and I do not want to spend the effort on getting the actual value
for your nonsense.

Since P=I^2R and power in your scenario is 100 W and the total resistance
is 50.001 Ohms, the current is 1.414 A.

Again since P=I^R and the antenna has a resistive impedance of 0.001 Ohms,
the power in the antenna is 0.002 W.

Do look up impedance matching and why it is used.




--
Jim Pennino

"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.


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?

On 10/27/2014 7:38 PM, gareth 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.


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?

You are right. Very little of the power reaching the end of that cable
will be radiated. Most will be reflected back down the cable toward the
source. When you add a 50 ohm termination you will still have very
little radiated. Nearly all of the power will be dissipated in the
resistor.

I think you have a very clear picture of what is happening.

--

Rick

rickman wrote:
On 10/27/2014 7:38 PM, gareth 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.


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?

You are right. Very little of the power reaching the end of that cable
will be radiated. Most will be reflected back down the cable toward the
source. When you add a 50 ohm termination you will still have very
little radiated. Nearly all of the power will be dissipated in the
resistor.

I think you have a very clear picture of what is happening.

Nope, it just shows he does not understand the concept of impedance
matching.



--
Jim Pennino

gareth 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.


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?

Assuming an antenna impedance of 0.001 Ohms, about 0.002 W.

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

About 0.002 W.

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

Yes.

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

About 99.998 W.

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

Nil.

Still no clue how antennas work, I see.



--
Jim Pennino

"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?

"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.