I have been informed my GRNDWAV3 program is in error - it calculates
the power input to a matched receiver to be 6dB greater than it ought
to be, or exactly 4 times the correct power input.

Most of my programs calculate results based on what I consider to be
fundamental reasoning. But GRNDWAV3 is one of the few where I have
adapted formulae from the text books or 'bibles'.

My informant is an Icelandic amateur who appears to know what he is
talking about and is mathematically very convincing. For various
resons, for the time being I propose to leave him out of this
discussion. That is, of course, if a discussion should evolve.

The problem fundamentally revolves around the gain of short vertical
antennas, both transmitting and receiving, above a perfect ground,
relative to isotropic. But for present purposes what an isotrope
actually is can be forgotten about. It exists only in one's
imagination.

Numbers cannot be avoided. So let's keep them as simple as possible by
starting with the MF standard of 1 Kilowatt, radiated from a short
vertical antenna above a perfect ground. Actual antenna height and
frequency don't matter.

According to the text books, the field strength from 1 Kw at 1
kilometre = 300 millivolts, which (according to the text books) is
correctly calculated by my program.

To calculate matched reciever input power from field strength it is
necessary to state vertical antenna height, frequency and radiation
resistance. Again choosing simple values -

Antenna height = 1 metre.
Frequency = 20 MHz.
Calculated radiation resistance = 1.758 ohms.
Matched receiver input resistance is also 1.758 ohms.

According to requirements antenna height is short compared with a
wavelength. I am confident that radiation resistance is correct at 20
MHz for a 1 metre vertical.

Antenna reactance is tuned out and disappears from the argument.

So we have a simple circuit consisting of a generator with a resistive
load of the same value, both equal to 1.758 ohms.

According to the text books (as confirmed by Roy) the generator
voltage is 300 millivolts. (A 1 metre high antenna with a field
strength of 300 mV per metre.)

The power available to the receiver is therefore -

Pr = Square( 0.3/2 ) divided by 1.758 = 12.8 milliwatts.

Which is the value calculated by my program although it does it in a
different way by not involving field strength. It calculates it more
directly from the 1 kW transmitter power and the antenna gains of a
pair of vertical Tx and Rx antennas relative to isotropic.

Nevertheless, I think my informant may be correct. That indeed my
program states receiver power input to be 4 times greater than what it
actually is. IMPORTANTLY, he says an NEC numerical program confirms
his own calculations.

NEC programs are not dependent on what a program user's ideas may be
about antenna gains relative to isotropic. They calculate directly
from fundamental metre-amps and volts.

I am presently out of touch with my informant. I do not know which
NEC program confirms his calculations.

I have recently asked Roy what is the voltage measured between the
bottom end of a 1 metre long vertical antenna and ground when the
field strength is 1 volt per metre. He says it is 1 volt and no doubt
the Bibles agree.

It is intriging, if the value should be only 0.5 volts then my program
would give the (suspected) correct answer to the simple question -

"What is the power input to a matched receiver using a 1 metre
vertical antenna, at 20 MHz, at a distance of 1 Km from a 1Kw
transmitter also using a short vertical antenna?" Short is less than
1/4-wavelength.

Is it 12.8 milliwatts, or is it 3.2 milliwatts?

Is there an NEC numerical program which will do the job? If there is
perhaps somebody could use it.

Most important, do I have to correct the program bug for the sake of 6
dB when the calculating uncertainty at long distances is plus or minus
10 or 15 dB ?
----
Reg, G4FGQ

Reg, G4FGQ wrote:
"He says it is 1 volt and no doubt the bibles agree."

OK. You have an antenna with some radiation resistance and a lossless
conjugate match to a load.

1/2 the antenna voltage which equals the volts per meter field strength
in your 1-meter wire, is dropped across the radiation resistance and the
other 0.5 volt appears across the receiver load. The radiation
resistance of the antenna becomes the Thevenin equivalent source
resistance of the generator feeding the receiver load..

The power lost to reradiation is 0.5 volt times the current in the
radiation resistance. The power delivered to the matched receiver load
is exactly the same.

Best regards, Richard Harrison, KB5WZI

Reg Edwards wrote:
I have been informed my GRNDWAV3 program is in error - it calculates
the power input to a matched receiver to be 6dB greater than it ought
to be, or exactly 4 times the correct power input.

Most of my programs calculate results based on what I consider to be
fundamental reasoning. But GRNDWAV3 is one of the few where I have
adapted formulae from the text books or 'bibles'.

My informant is an Icelandic amateur who appears to know what he is
talking about and is mathematically very convincing. For various
resons, for the time being I propose to leave him out of this
discussion. That is, of course, if a discussion should evolve.

The problem fundamentally revolves around the gain of short vertical
antennas, both transmitting and receiving, above a perfect ground,
relative to isotropic. But for present purposes what an isotrope
actually is can be forgotten about. It exists only in one's
imagination.

Numbers cannot be avoided. So let's keep them as simple as possible by
starting with the MF standard of 1 Kilowatt, radiated from a short
vertical antenna above a perfect ground. Actual antenna height and
frequency don't matter.

According to the text books, the field strength from 1 Kw at 1
kilometre = 300 millivolts, which (according to the text books) is
correctly calculated by my program.

To calculate matched reciever input power from field strength it is
necessary to state vertical antenna height, frequency and radiation
resistance. Again choosing simple values -

Antenna height = 1 metre.
Frequency = 20 MHz.
Calculated radiation resistance = 1.758 ohms.
Matched receiver input resistance is also 1.758 ohms.

According to requirements antenna height is short compared with a
wavelength. I am confident that radiation resistance is correct at 20
MHz for a 1 metre vertical.

Antenna reactance is tuned out and disappears from the argument.

So we have a simple circuit consisting of a generator with a resistive
load of the same value, both equal to 1.758 ohms.

According to the text books (as confirmed by Roy) the generator
voltage is 300 millivolts. (A 1 metre high antenna with a field
strength of 300 mV per metre.)

The power available to the receiver is therefore -

Pr = Square( 0.3/2 ) divided by 1.758 = 12.8 milliwatts.

Which is the value calculated by my program although it does it in a
different way by not involving field strength. It calculates it more
directly from the 1 kW transmitter power and the antenna gains of a
pair of vertical Tx and Rx antennas relative to isotropic.

Nevertheless, I think my informant may be correct. That indeed my
program states receiver power input to be 4 times greater than what it
actually is. IMPORTANTLY, he says an NEC numerical program confirms
his own calculations.

NEC programs are not dependent on what a program user's ideas may be
about antenna gains relative to isotropic. They calculate directly
from fundamental metre-amps and volts.

I am presently out of touch with my informant. I do not know which
NEC program confirms his calculations.

I have recently asked Roy what is the voltage measured between the
bottom end of a 1 metre long vertical antenna and ground when the
field strength is 1 volt per metre. He says it is 1 volt and no doubt
the Bibles agree.

It is intriging, if the value should be only 0.5 volts then my program
would give the (suspected) correct answer to the simple question -

"What is the power input to a matched receiver using a 1 metre
vertical antenna, at 20 MHz, at a distance of 1 Km from a 1Kw
transmitter also using a short vertical antenna?" Short is less than
1/4-wavelength.

Is it 12.8 milliwatts, or is it 3.2 milliwatts?

Is there an NEC numerical program which will do the job? If there is
perhaps somebody could use it.

Most important, do I have to correct the program bug for the sake of 6
dB when the calculating uncertainty at long distances is plus or minus
10 or 15 dB ?
----
Reg, G4FGQ


How was your Icelandic amateur doing his NEC calculations? If he
calculated the voltage that would generate 1kW into a load, then excited
his transmit antenna with a matched generator of that source impedance
-- which would drop the power by a factor of four. If he did that and
you did your calculations with 1kW going _to_ the transmit antenna that
would be your source of error.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

"Tim Wescott" wrote in message
...
Reg Edwards wrote:
I have been informed my GRNDWAV3 program is in error - it
calculates
the power input to a matched receiver to be 6dB greater than it
ought
to be, or exactly 4 times the correct power input.

Most of my programs calculate results based on what I consider to
be
fundamental reasoning. But GRNDWAV3 is one of the few where I have
adapted formulae from the text books or 'bibles'.

My informant is an Icelandic amateur who appears to know what he
is
talking about and is mathematically very convincing. For various
resons, for the time being I propose to leave him out of this
discussion. That is, of course, if a discussion should evolve.

The problem fundamentally revolves around the gain of short
vertical
antennas, both transmitting and receiving, above a perfect ground,
relative to isotropic. But for present purposes what an isotrope
actually is can be forgotten about. It exists only in one's
imagination.

Numbers cannot be avoided. So let's keep them as simple as
possible by
starting with the MF standard of 1 Kilowatt, radiated from a short
vertical antenna above a perfect ground. Actual antenna height and
frequency don't matter.

According to the text books, the field strength from 1 Kw at 1
kilometre = 300 millivolts, which (according to the text books) is
correctly calculated by my program.

To calculate matched reciever input power from field strength it
is
necessary to state vertical antenna height, frequency and
radiation
resistance. Again choosing simple values -

Antenna height = 1 metre.
Frequency = 20 MHz.
Calculated radiation resistance = 1.758 ohms.
Matched receiver input resistance is also 1.758 ohms.

According to requirements antenna height is short compared with a
wavelength. I am confident that radiation resistance is correct at
20
MHz for a 1 metre vertical.

Antenna reactance is tuned out and disappears from the argument.

So we have a simple circuit consisting of a generator with a
resistive
load of the same value, both equal to 1.758 ohms.

According to the text books (as confirmed by Roy) the generator
voltage is 300 millivolts. (A 1 metre high antenna with a field
strength of 300 mV per metre.)

The power available to the receiver is therefore -

Pr = Square( 0.3/2 ) divided by 1.758 = 12.8 milliwatts.

Which is the value calculated by my program although it does it in
a
different way by not involving field strength. It calculates it
more
directly from the 1 kW transmitter power and the antenna gains of
a
pair of vertical Tx and Rx antennas relative to isotropic.

Nevertheless, I think my informant may be correct. That indeed my
program states receiver power input to be 4 times greater than
what it
actually is. IMPORTANTLY, he says an NEC numerical program
confirms
his own calculations.

NEC programs are not dependent on what a program user's ideas may
be
about antenna gains relative to isotropic. They calculate directly
from fundamental metre-amps and volts.

I am presently out of touch with my informant. I do not know
which
NEC program confirms his calculations.

I have recently asked Roy what is the voltage measured between the
bottom end of a 1 metre long vertical antenna and ground when the
field strength is 1 volt per metre. He says it is 1 volt and no
doubt
the Bibles agree.

It is intriging, if the value should be only 0.5 volts then my
program
would give the (suspected) correct answer to the simple
uestion -

"What is the power input to a matched receiver using a 1 metre
vertical antenna, at 20 MHz, at a distance of 1 Km from a 1Kw
transmitter also using a short vertical antenna?" Short is less
than
1/4-wavelength.

Is it 12.8 milliwatts, or is it 3.2 milliwatts?

Is there an NEC numerical program which will do the job? If there
is
perhaps somebody could use it.

Most important, do I have to correct the program bug for the sake
of 6
dB when the calculating uncertainty at long distances is plus or
minus
10 or 15 dB ?
----
Reg, G4FGQ


How was your Icelandic amateur doing his NEC calculations? If he
calculated the voltage that would generate 1kW into a load, then
excited
his transmit antenna with a matched generator of that source
impedance
-- which would drop the power by a factor of four. If he did that
and
you did your calculations with 1kW going _to_ the transmit antenna
that
would be your source of error.

=====================================
Tim,

I think the error, if there is one, is most likely at the receiving
end.

The RADIATED power is 1000 watts.
It doesn't matter how it got into the ether from the transmitter.

But you have made me think again about what should be done with
transmitting antenna gain.
----
Reg.

On Thu, 8 Dec 2005 21:44:07 +0000 (UTC), "Reg Edwards"
wrote:

The problem fundamentally revolves around the gain of short vertical
antennas, both transmitting and receiving,

Hi Reggie,

This is not what you proposed the first time through this exercise. As
your original was stated in terms of a field of 1V/M there is no need
to elaborate on about the radiator.

Numbers cannot be avoided. So let's keep them as simple as possible by
starting with the MF standard of 1 Kilowatt, radiated from a short
vertical antenna above a perfect ground. Actual antenna height and
frequency don't matter.

Then its mention at being both MF (MW?) standard, and being a short
vertical is in distinct conflict. Again, there is nothing inherent to
the problem short of demonstrable evidence in the field, which returns
us to the classic FCC ground wave charts for MF (MW?).

According to the text books, the field strength from 1 Kw at 1
kilometre = 300 millivolts, which (according to the text books) is
correctly calculated by my program.

Are you claiming your program computes field V/M or power delivered by
a field of V/M?

To calculate matched reciever input power from field strength

Here again is a new specification to your original query. Matched? I
would have suspected so, but being unmatched or being mismatched
brings a spectrum of answers that span to more than 1V/M down to
microV/M.

it is
necessary to state vertical antenna height, frequency and radiation
resistance. Again choosing simple values -

Antenna height = 1 metre.
Frequency = 20 MHz.
Calculated radiation resistance = 1.758 ohms.
Matched receiver input resistance is also 1.758 ohms.

Even more elaboration that goes beyond the original. I suppose for
the purposes of discussion it works, but it goes well beyond the
premise of classic MF (MW?) work.

According to requirements antenna height is short compared with a
wavelength. I am confident that radiation resistance is correct at 20
MHz for a 1 metre vertical.

When we have theoretical work supported by real field data in the MW
band, why take up a frequency outside that?

Antenna reactance is tuned out and disappears from the argument.

This is to be expected, but was never actually expressed anywhere.

So we have a simple circuit consisting of a generator with a resistive
load of the same value, both equal to 1.758 ohms.

The generator's Z is immaterial to the discussion if you have a
defined field.

According to the text books (as confirmed by Roy) the generator
voltage is 300 millivolts. (A 1 metre high antenna with a field
strength of 300 mV per metre.)

I have obtained comparable (though not exact) results for real (not
perfect) grounds to compare against actual (not theoretical) data.
Those not comparable have come from mismatched 1M high antennas
against various loads (already discussed above).

The power available to the receiver is therefore -

Pr = Square( 0.3/2 ) divided by 1.758 = 12.8 milliwatts.

Where'd the 2 of the ( 0.3/2 ) come from?

Which is the value calculated by my program

Nevertheless, I think my informant may be correct.

It is quite simple to fool experts. They are by far easier targets
than the naive who ask for the missing work (like this fool here who
wonders just what your correspondent offered to convince you).

Is it 12.8 milliwatts, or is it 3.2 milliwatts?

EZNEC computes it at E²/R

73's
Richard Clark, KB7QHC

Hi Richard,

You must have got out of bed the wrong side this morning. ;o)
----
Yours, Punchinello

Sorry, I forgot "per metre".

I should have said -

"According to the text books, the field strength from 1 Kw at 1
kilometre = 300 millivolts per metre."
----
Reg.

The answer is 3 mW.

Any version of EZNEC can be used to do this calculation. The demo
program will yield slightly less accurate results because of the limited
number of segments(*).

I modeled two vertical wires, 1 meter high and 1 mm diameter, spaced 1
km apart, at 20 MHz, over perfect ground. The reported feedpoint
impedance varies with segmentation, from 1.988 - j952.3 ohms at 10
segments/wire to 1.72 - j882 ohms at 100 segments/wire. Accuracy is
likely to degrade with a larger number of segments, since even 100
results in segment length/diameter ratio less than NEC recommendations.
I used 100 segments/wire for the test.

One of the choices in EZNEC of far field strength reporting is in V/m at
1 kW input and 1 km distance. For this antenna, EZNEC reports 300.8 V/m
(RMS) at ground level.

EZNEC also permits setting a fixed power input, so this was set to 1 kW.
The resulting source voltage and current are 21270 V. and 24.12 A.
respectively.

A load of 1.72 + j882 ohms was placed at the base of the second
vertical. EZNEC reports a power of 3.234 mW being dissipated in this load.

Care has to be used when analyzing the current induced in one antenna by
another which is distant using numerical calculations. Errors can occur
due to truncation and other causes when the ratio of distances between
the two antennas is great relative to the segment lengths or to segment
distances within one of the antennas. However, EZNEC gets virtually
identical results when using mixed and double precision NEC-2
calculating engines, which indicates that the limit hasn't been reached
and that numerical problems aren't occurring. (Another check which can
be done is to reduce the distance between antennas by a factor of two.
The power in the load resistance should increase by a factor of four.)

Another critical matter is the setting of the load reactance. The
reactance is many times larger than the resistance, so a slight error in
setting its value will result in a large difference in load current and
therefore load dissipation. For example, if the segmentation is changed
from 100 to 50 segments/wire and no other change is made to the model,
the reported load power becomes 0.3917 watts. The reason is that the
reported source impedance is now 1.756 - j891.4 ohms, while the load is
still 1.72 + j882 ohms. Changing the load to the proper conjugately
matched value of 1.756 + 891.4 ohms returns the load power to the
correct value of 3.24 mW.

All given, I'd trust the reported load power to be easily within 10% of
the theoretically correct value.

(*) Results for 10 segments/wire are 1.988 - j952.3 ohms for the source
impedance, 300.71 V/m field strength at 1 km for 1 kW, and 3.24 mW in a
conjugately matched load impedance in the distant vertical.

Roy Lewallen, W7EL

On Fri, 9 Dec 2005 00:04:38 +0000 (UTC), "Reg Edwards"
wrote:
You must have got out of bed the wrong side this morning. ;o)

Ah hah,

Another troll using the Keflavik proxy, eh wot?

Yes, we should all beware a straight question from a crooked source.

73's
Richard Clark, KB7QHC

Roy Lewallen wrote:
. . .
Another critical matter is the setting of the load reactance. The
reactance is many times larger than the resistance, so a slight error in
setting its value will result in a large difference in load current and
therefore load dissipation. For example, if the segmentation is changed
from 100 to 50 segments/wire and no other change is made to the model,
the reported load power becomes 0.3917 watts. . . .

Correction: That should be 0.3917 mW. The error doesn't alter the
conclusion.

Roy Lewallen, W7EL