[ Copied from another newsgroup.]

While I'm in the mood, the following little calculation may assist in
understanding what goes on in an antenna. Suppose we wish to know the
radiating efficiency of a resonant half-wave dipole fed at its center.


When at exact resonance we are concerned only with pure resistances.


For calculating convenience, we assume the radiation resistance, Rrad, is
uniformly distributed along the length of the wire and is 140 ohms which has
been calculated from its dimensions. It only has two - Length and
Diameter. But for a half-wave dipole it is always about 140 ohms. Wire
diameter has a relatively small effect on Rrad.


Frequency, F = 10 MHz.
One wavelength = 300/F = 30 metres.
Overall dipole length L = 15 metres.
Copper wire diameter D = 2mm.


Calculate Rloss, the end-to-end wire loss resistance -


Rloss = 0.0833 * Sqrt( F ) * Length / Diameter ohms, which takes skin
effect with copper wire into account.


Rloss = 1.976 ohms.

Insofar as the current flow is concerned, the total resistance in the
antenna wire is 140 + 1.976 = 141.976 ohms. But only Sqr( Amps ) * Rloss
watts is lost.


So antenna efficiency = (Power radiated) / (Power input) = Rrad / ( Rrad +
Rloss )


Effncy = 140 / ( 140 + 1.976 ) times 100 percent.


Effncy = 98.6 percent.

Note that we have taken care to place both the radiation and loss
resistances at the same place in the system so that they can be properly
compared. It is usual to locate the radiation resistance at the dipole
centre as one lump when it then becomes 70 ohms. But a slightly different
efficiency calculation is then needed to provide the same correct answer.


Almost certainly there will be an appreciably greater loss in the
transmission line to the transmitter and in the coupling/impedance
transforming coil and capacitor networks.


If you are interested in numerical quantities without tedious arithmetic,
download in a few seconds one or two simple programs from the following
website and run immediately. Amuse yourself.
----
.................................................. ...........
Regards from Reg, G4FGQ
For Free Radio Design Software go to
http://www.btinternet.com/~g4fgq.regp
.................................................. ..........

Reg:

[snip]
For calculating convenience, we assume the radiation resistance, Rrad, is
uniformly distributed along the length of the wire and is 140 ohms which
has
been calculated from its dimensions. It only has two - Length and
Diameter. But for a half-wave dipole it is always about 140 ohms. Wire
diameter has a relatively small effect on Rrad.
[snip]

Reg, in your model, is your *assumption* "for calculating convenience" that
radiation
resistance is uniformly distributed along the antenna structure, i.e. the
transmission line
that represents the antenna in your model, supported by any theory or is it
just a
mathematical *fit* to the data?

For example, one could *assume* literally any analytic distribution of
radiation resistance
along an antenna's length, for instance sinusoidal, catenary, exponential,
triangular, etc...
and come up with a value/function for that particular distribution that has
the equivalent
effect of a lumped value placed at the antenna feedpoint. What is so unique
about uniform?

Why do you think *uniform* is any better than any other distribution of
Rrad?

I have no axe to grind here, just curiosity...

Best Regards for the New Year.

--
Peter K1PO
Indialantic By-the-Sea, FL.

Peter,

I am somewhat surprised to receive such questions from your good self. They
are not so far from the realms of Ohm's Law as to cause YOU any
difficulties. Perhaps after the festivities you are feeling too lazy to
satisfy your own curiosity by exercising your brain cells. ;o)

You must be aware, even without thinking about it, a lumped radiation
resistance must always be associated with a definite location on an antenna
at which the current is known. This by no means need be at the feedpoint.
But I guess this is the first occasion on which you have been confronted
with the *distributed* variety and have been brought to a sudden dead stop.

Let's stay with the well-known resonant 1/2-wave dipole. The objective is
to directly compare radiation resistance with wire loss resistance. To do
this means the same current must flow through both just as if they were in
series with each other.

{ Many people are familiar with the simple equation, efficiency = Rrad /
( Rrad + Rloss ) and state it whenever an appropriate occasion arises. It
sounds very learned of course. But in the whole of North America I venture
to say hardly a single radio amateur knows from where Rloss and Rrad can be
obtained (except perhaps ground loss with verticals) and what its value is.
It follows that few have ever used the equation presented in Handbook
articles, etc. }

We have a choice. 1. Lump both the radiation resistance and conductor
resistance together at one point after transforming from the distributed to
lumped value of wire loss.

Or 2, leave the wire resistance where it is and distribute the radiation
resistance along the wire. We have no choice about the type of
istribution - it must be the same as the wire resistance is distributed -
i.e., uniformly.

Whatever we do we cannot avoid transforming from a lumped to distributed
resistance value, or vice-versa. Electrical engineers do it all the time.
In the case of a dipole there are several ways. But its a simple process
and the result is amazingly even more simple.

I prefer to begin with the accurate assumption of a sinewave distribution of
current along the dipole wire with the maximum of 1 amp at the dipole
centre. Then integrate P = I squared R from one end of the wire to the
other to find the total power dissipated in the wire.

The equivalent lump of resistance located at the centre (where 1 amp flows)
turns out to be exactly half of uniformly distributed end-to-end resistance
of the wire. In fact, that's exactly how the radiation resistance of the
usual 70-ohm lump got itself into a dipole's feedpoint. It is exactly half
of 140 ohms. If radiation resistance itself had any say in the matter I am
sure it would prefer to be nicely spread along the length of the wire
instead of being stuck in a lump next to the feedpoint.

If the end-to-end wire loss resistance is R ohms then the ficticious
equivalent lump at the centre feedpoint is exactly R/2 ohms. So easy to
remember, eh?

Another way of obtaining exactly the same result is to calculate the input
impedance of a 1/4-wave, open-circuit, transmission line, which of course is
the same as half of a half-wave dipole. It even has a 1/4-sinewave current
distribution along its length. The input resistance at resonance is always
half of the conductor loss resistance. With a good impedance bridge this can
be measured to keep Roy happy.

In fact, it is the pair of 1/4-wave, open-circuit, single-wire lines
constituting the dipole which transform the uniformly distributed wire loss
resistance to the equivalent lumped 1/2-value input resistances as measured
at the dipole centre. And, of course, the antenna performs exactly the same
transformation on an antenna's uniformly distributed radiation resistance.
I sometimes feel sorry for things which find themselves securely locked in,
constrained for ever to obey the irresistible laws of nature, helpless to do
othewise, for ever.

See how the interlocking bits of the jig-saw puzzle now fit very nicely
together.

Your general question - yes it would be possible to 'assume' any arbitrary
mathematical distribution of radiation or loss resistance and then find an
equivalent lumped value which would radiate/dissipate the same power when
located at a particular current point. But it would not be of any practical
use - it would never correspond to an actual antenna. When calculating
efficiency of wire antennas it seems only a uniform distribution of
resistance is of any use. An investigator has no choice in the matter.

Calculating the efficiency of coil loaded antennas gets complicated. The
current distributions of the upper and lower sections are different and so
are their efficiencies. But efficiencies are so high in the conductors
themselves ball-park guesses are good enough. However it is still necessary
to transform various effects, including those due to the coil, to the common
base feedpoint in order to calculate input impedance.
---
Best Wishes, Reg, G4FGQ

===================================

Reg:

[snip]
For calculating convenience, we assume the radiation resistance, Rrad,
is
uniformly distributed along the length of the wire and is 140 ohms which
has
been calculated from its dimensions. It only has two - Length and
Diameter. But for a half-wave dipole it is always about 140 ohms. Wire
diameter has a relatively small effect on Rrad.
[snip]

Reg, in your model, is your *assumption* "for calculating convenience"
that
radiation
resistance is uniformly distributed along the antenna structure, i.e. the
transmission line
that represents the antenna in your model, supported by any theory or is
it
just a
mathematical *fit* to the data?

For example, one could *assume* literally any analytic distribution of
radiation resistance
along an antenna's length, for instance sinusoidal, catenary, exponential,
triangular, etc...
and come up with a value/function for that particular distribution that
has
the equivalent
effect of a lumped value placed at the antenna feedpoint. What is so
unique
about uniform?

Why do you think *uniform* is any better than any other distribution of
Rrad?

I have no axe to grind here, just curiosity...

Best Regards for the New Year.

--
Peter K1PO
Indialantic By-the-Sea, FL.

Peter O. Brackett wrote:
Why do you think *uniform* is any better than any other distribution of
Rrad?

Here's a quote from _Antennas_, by Kraus & Marhefka: "... the radiation
resistance may be thought of as a virtual resistance that does not exist
physically but is a quantity coupling the antenna to distant regions of
space via a virtual transmission line."
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil, W5DXP wrote:
"Here`s a quote from _Antennas_by Kraus&Marthelka: "... the radiation
resistance may be thought of as a virtual resistance that does not exist
physically but is a quantity coupling the antenna to distant regions of
space via a virtual transmission line."

A problem comparing radiation to a transmission line is that everytime
line length increases a fixed amount, so does the loss. The loss is so
many db per mile. After a radio wave travels a million miles, the next
million miles only produces an additional 6 db loss. Thus, we are
looking at radio pictures from Mars. We couldn`t get them by the lowest
loss coax from Mars without distributed repeaters.

I can`t find my 1955 edition of Terman but as I recall he says that
radiation resistance is equal to the resistor, which if placed in series
with the antenna, would consume the same power that the antenna does.
Radiation resistance is a useful fiction. Radiation resistance is
usually the resistance at an antenna`s maximum current point. Fictional
or not, it can be measured with an impedance bridge.

My 1937 edition of Terman says:
"---fields produced by the antenna currents induce back voltages that
are equivalent to an impedance against which the applied voltage can be
considered as acting."

Terman also says: "In addition to the radiated energy, energy is also
lost in the antenna system as a result of wire and ground resistance,
corona, eddy currents induced in neighboring masts, guy wires and other
conductors, and dielectric losses arising from such imperfect
dielectrics as trees and insulators, located in the field of the
antenna. These losses can be represented in the same way as the radiated
energy, i.e., by a resistance which inserted in series with the antenna
will consume the same amount of power as that actually dissipated in
these various ways. The total antenna resistance is the sum, Rr + Rl, of
the radiation resistance Rr and the loss resistance Rl, and determines
the amount of energy that must be supplied to the antenna to produce a
given current.

The efficiency of the antenna as a radiator is the ratio Rr / (Rr + Rl)
of radiation to total resistance. This represents the fraction of the
total energy supplied to the antenna which is converted into radio
waves."

Terman also says:
"The radiated field (epsilon) varies directly as the current I, the
frequency f, the doublet length (delta)l, and the cosine of the angle of
elevation, and is inversely proportional to the distance d."

The strength of the radiation from an antenna at a point P is the sum of
the strengths of the fields of its elemental pieces carrying a current
I.

Current is not uniform throughout the usual antenna and neither is the
antenna`s impedance. The current is zero at the open ends of an antenna,
near zero at exact multiples of a half wave length distant from the open
end, while the current is maximum at points that are odd quarter wave
lengths distant from the open ends.

Best regards, Richard Harrison, KB5WZI

Very, very good Reg. If you can get them to understand
what you have said, it would be a cakewalk to understand
the underpinnings of my antennas. My aproach of explaining
what you have said is to refer to 'complex circuitry',
where in general use this is used to SIMPLIFY circuits.
If one understood the basics of what you are pointing out
then one could easily understand how one can translate
lumped circuits to circuits that RADIATE in an efficient manner.
It should not be difficult to understand that a matching circuit
which is desirable in a lot of cases, is a circuit of lumped items.
There is absolutely no reason whatsoever to prevent one from
transforming
the lumped items to radiating circuits which not only radiate but also
provide the same impedance to a transmitter that the original matching
unit supplied.
Also very importantly it shows how little the frequency dominates
the antenna size
Thanks a bunch for your posting and I wish you luck in your
education pursuit in the area that I met failure
Cheers
Art.


"Reg Edwards" wrote in message ...
Peter,

I am somewhat surprised to receive such questions from your good self. They
are not so far from the realms of Ohm's Law as to cause YOU any
difficulties. Perhaps after the festivities you are feeling too lazy to
satisfy your own curiosity by exercising your brain cells. ;o)

You must be aware, even without thinking about it, a lumped radiation
resistance must always be associated with a definite location on an antenna
at which the current is known. This by no means need be at the feedpoint.
But I guess this is the first occasion on which you have been confronted
with the *distributed* variety and have been brought to a sudden dead stop.

Let's stay with the well-known resonant 1/2-wave dipole. The objective is
to directly compare radiation resistance with wire loss resistance. To do
this means the same current must flow through both just as if they were in
series with each other.

{ Many people are familiar with the simple equation, efficiency = Rrad /
( Rrad + Rloss ) and state it whenever an appropriate occasion arises. It
sounds very learned of course. But in the whole of North America I venture
to say hardly a single radio amateur knows from where Rloss and Rrad can be
obtained (except perhaps ground loss with verticals) and what its value is.
It follows that few have ever used the equation presented in Handbook
articles, etc. }

We have a choice. 1. Lump both the radiation resistance and conductor
resistance together at one point after transforming from the distributed to
lumped value of wire loss.

Or 2, leave the wire resistance where it is and distribute the radiation
resistance along the wire. We have no choice about the type of
istribution - it must be the same as the wire resistance is distributed -
i.e., uniformly.

Whatever we do we cannot avoid transforming from a lumped to distributed
resistance value, or vice-versa. Electrical engineers do it all the time.
In the case of a dipole there are several ways. But its a simple process
and the result is amazingly even more simple.

I prefer to begin with the accurate assumption of a sinewave distribution of
current along the dipole wire with the maximum of 1 amp at the dipole
centre. Then integrate P = I squared R from one end of the wire to the
other to find the total power dissipated in the wire.

The equivalent lump of resistance located at the centre (where 1 amp flows)
turns out to be exactly half of uniformly distributed end-to-end resistance
of the wire. In fact, that's exactly how the radiation resistance of the
usual 70-ohm lump got itself into a dipole's feedpoint. It is exactly half
of 140 ohms. If radiation resistance itself had any say in the matter I am
sure it would prefer to be nicely spread along the length of the wire
instead of being stuck in a lump next to the feedpoint.

If the end-to-end wire loss resistance is R ohms then the ficticious
equivalent lump at the centre feedpoint is exactly R/2 ohms. So easy to
remember, eh?

Another way of obtaining exactly the same result is to calculate the input
impedance of a 1/4-wave, open-circuit, transmission line, which of course is
the same as half of a half-wave dipole. It even has a 1/4-sinewave current
distribution along its length. The input resistance at resonance is always
half of the conductor loss resistance. With a good impedance bridge this can
be measured to keep Roy happy.

In fact, it is the pair of 1/4-wave, open-circuit, single-wire lines
constituting the dipole which transform the uniformly distributed wire loss
resistance to the equivalent lumped 1/2-value input resistances as measured
at the dipole centre. And, of course, the antenna performs exactly the same
transformation on an antenna's uniformly distributed radiation resistance.
I sometimes feel sorry for things which find themselves securely locked in,
constrained for ever to obey the irresistible laws of nature, helpless to do
othewise, for ever.

See how the interlocking bits of the jig-saw puzzle now fit very nicely
together.

Your general question - yes it would be possible to 'assume' any arbitrary
mathematical distribution of radiation or loss resistance and then find an
equivalent lumped value which would radiate/dissipate the same power when
located at a particular current point. But it would not be of any practical
use - it would never correspond to an actual antenna. When calculating
efficiency of wire antennas it seems only a uniform distribution of
resistance is of any use. An investigator has no choice in the matter.

Calculating the efficiency of coil loaded antennas gets complicated. The
current distributions of the upper and lower sections are different and so
are their efficiencies. But efficiencies are so high in the conductors
themselves ball-park guesses are good enough. However it is still necessary
to transform various effects, including those due to the coil, to the common
base feedpoint in order to calculate input impedance.
---
Best Wishes, Reg, G4FGQ

===================================

Reg:

[snip]
For calculating convenience, we assume the radiation resistance, Rrad,
is
uniformly distributed along the length of the wire and is 140 ohms which
has
been calculated from its dimensions. It only has two - Length and
Diameter. But for a half-wave dipole it is always about 140 ohms. Wire
diameter has a relatively small effect on Rrad.
[snip]

Reg, in your model, is your *assumption* "for calculating convenience"
that
radiation
resistance is uniformly distributed along the antenna structure, i.e. the
transmission line
that represents the antenna in your model, supported by any theory or is
it
just a
mathematical *fit* to the data?

For example, one could *assume* literally any analytic distribution of
radiation resistance
along an antenna's length, for instance sinusoidal, catenary, exponential,
triangular, etc...
and come up with a value/function for that particular distribution that
has
the equivalent
effect of a lumped value placed at the antenna feedpoint. What is so
unique
about uniform?

Why do you think *uniform* is any better than any other distribution of
Rrad?

I have no axe to grind here, just curiosity...

Best Regards for the New Year.

--
Peter K1PO
Indialantic By-the-Sea, FL.

Reg:

[snip]
Perhaps after the festivities you are feeling too lazy to
satisfy your own curiosity by exercising your brain cells. ;o)
[snip]

Ahem... well I do admit to imbibing during the Holiday, but I feel that at
least
20% of my brain cells are still intact which should enable me to pass the
next generation of ham radio exams with no problem. :-)


[snip]
But I guess this is the first occasion on which you have been confronted
with the *distributed* variety and have been brought to a sudden dead
stop.
[snip]

Well yes and no!

Even tho,. for professional reasons, I have extensive transmission line
modelling
software [self-developed] which supports extremes of complex Zo and
distributed
losses with various loss distributions along the lines, I have never used
these
computer codes/algorithms to simulate antennas.

[My professional applications of these codes, written in Fortran, have been
for broadband
digital subscriber loop, DSL, BRA ISDN and cable modem transmissions over
telco local
loops. i.e. upwards of 1000 to18,000 feet of twisted pairs of mixed guages
and dielectrics,
with bridged taps etc. These codes allow for empirical fits to primary
parameters, R, L, C and G
as functions of frequency and other effects, etc... I had posted on this NG
some of the models
developed by several contributors to the ANSI T1E1.4 Standards Committee
over the past
few years sometime in the last year or so if you recall.]

Clearly such software/algorithms which are sort of like finite element
analysis methods breaking
the lines into incremental sections and summing the results, etc... and can
also be used to simulate
the driving point impedances and losses, both disipative and radiative, of
antennas as you suggest.

Until your posting I had never fully thought through what the distribution
of radiative losses
on antenna structures should be...

[snip]
Or 2, leave the wire resistance where it is and distribute the radiation
resistance along the wire. We have no choice about the type of
istribution - it must be the same as the wire resistance is
istributed -
i.e., uniformly.
:
:
Whatever we do we cannot avoid transforming from a lumped to distributed
resistance value, or vice-versa. Electrical engineers do it all the time.
In the case of a dipole there are several ways. But its a simple process
and the result is amazingly even more simple.
:
:
The equivalent lump of resistance located at the centre (where 1 amp
flows)
turns out to be exactly half of uniformly distributed end-to-end
resistance
of the wire. In fact, that's exactly how the radiation resistance of the
usual 70-ohm lump got itself into a dipole's feedpoint. It is exactly
half
of 140 ohms. If radiation resistance itself had any say in the matter I
am
sure it would prefer to be nicely spread along the length of the wire
instead of being stuck in a lump next to the feedpoint.

If the end-to-end wire loss resistance is R ohms then the ficticious
equivalent lump at the centre feedpoint is exactly R/2 ohms. So easy to
remember, eh?
[snip]

Yes it sure is!

[snip]
In fact, it is the pair of 1/4-wave, open-circuit, single-wire lines
constituting the dipole which transform the uniformly distributed wire
loss
resistance to the equivalent lumped 1/2-value input resistances as
measured
at the dipole centre. And, of course, the antenna performs exactly the
same
transformation on an antenna's uniformly distributed radiation resistance.
I sometimes feel sorry for things which find themselves securely locked
in,
constrained for ever to obey the irresistible laws of nature, helpless to
do
othewise, for ever.

See how the interlocking bits of the jig-saw puzzle now fit very nicely
together.
[snip]

Linear distribution...

Yes, now with your simple, yet very clear explanation, I now see that,
thanks!

[snip]
\ use - it would never correspond to an actual antenna. When calculating
efficiency of wire antennas it seems only a uniform distribution of
resistance is of any use. An investigator has no choice in the matter.
[snip]

Hmmm... I'm just thinking... that may not always be the case!

What about certain kinds of travelling wave antennas. i.e. a V-beam,
or a rhombic, etc... which are transmission lines with an ever changing
spacing between the elements. Surely the radiation resistance along such an
antenna/transmission line is not distributed uniformly even tho the
dissipative
losses are!

Thanks again for your lucid reply, I am indebted to you for refreshing
some of my *besotted* brain cells... hmmm, I wonder is it the
reds or the whites that cause most of the brain cell damage?

I'm gonna go try some of my homebrew transmission line software on
some antenna problems and see how it does...

Best Regards for the New Year.

--
Peter K1PO
Indialantic By-the-Sea, FL

"Art Unwin KB9MZ" wrote
Very, very good Reg.

==========================

Art, nice to hear from a representative of the few who agree with what is
the bleeding obvious.

I am aware of your long outstanding problems about convincing folks of the
properties of your loop-coupled antenna proposals.

But I am too exhausted and too long-in-the-tooth to take part in the (by
far) unecessarily convoluted arguments.

Try KISS. Provide a precise, unambiguous picture of all dimensions and
submit it to a program capable of analysing it - if you can find one. I am
unable to provide any assistance myself in that direction.

May you and yours enjoy life in 2004 to the full.
----
Yours, Reg, G4FGQ.

"Reg Edwards" wrote in message
...
"Art Unwin KB9MZ" wrote
Very, very good Reg.

==========================

Art, nice to hear from a representative of the few who agree with what is
the bleeding obvious.

I am aware of your long outstanding problems about convincing folks of the
properties of your loop-coupled antenna proposals.

But I am too exhausted and too long-in-the-tooth to take part in the (by
far) unecessarily convoluted arguments.

Try KISS. Provide a precise, unambiguous picture of all dimensions and
submit it to a program capable of analysing it - if you can find one.

Reg, no need for help as it is all completed with success.
I may add. I used AO PRO to do the final wrap up as well as making the
antennas which in uncompromising fashion proved what you are stating but
what other people have been unable to understand .
And I have made many different antennas of this family.!
I might add that with shorting either capacitor one can change a "T match
style antenna "to other forms that provide for high or low impedance at the
antenna feed point. to meet requirements of the transmitter.as well as
providing a ' loss less' interface, a subject that has been bandied around
for years but in isolation.
I am sure glad however, to see a dissertation such as yours that was
unsolicitated even tho it may finish up as a 'Plonk" on this side of the
Pond as many times the obvious is ignored until it apears in a book
Hopefully you will able to withstand the junk that will now be thrown at
you for stating such an outrageous thing.
Cheers
Art


I am
unable to provide any assistance myself in that direction.

May you and yours enjoy life in 2004 to the full.
----
Yours, Reg, G4FGQ.

Peter,

To satisfy yourself that a half-wave dipole automatically transforms
end-to-end wire resistance to an equivalent lumped resistance of half its
value located at the dipole centre, use program RJELINE3. It takes only a
few seconds.

Enter F = 10 MHz, Open-wire line length = 7.5 metres = 1/4-wave.

As everybody knows a 1/4-wavelength of line (a half dipole), behaves as an
impedance transformer.

Any value Zo of open wire line will do. But try Zo around 500 ohms with thin
wire such as 0.2mm diameter.

Terminate the line with 99999999 + j99999999 ohms, ie., open circuit just
like the dipole ends.

Loop-ohms per metre of the wire is one of the computed results.

Another computed result is exact line length in wavelengths.

Vary line length until it is exactly 1/4 wavelengths.

The input impedance of the 1/4-wave length of open-circuited line is also
calculated and displayed.

It will be found that at exact resonance (vary length or frequency very
finely) the input impedance of the line will be a pure resistance ( jXin =
0) equal to half of the of the line end-to-end wire resistance.

It is obvious exactly the same transformation occurs when the wire
resistance is replaced by a uniformly distributed radiation resistance.

If your own programs significantly disagree then consign them to the junk
box.

As you may have noticed I never support my stuff by citing the usual old
wives. Never come across, even heard of most of 'em. There are no references
except my tattered note books. I came across various useful relationship
around 1960 when researching into methods of locating faults on oceanic
phone cables. But I daresay Heaviside preceded me. I dug up much information
and designed fault locating and other test equipment but very little was
published beyond contract manufacturing information. There were two articles
in the house engineering journal. I worked alone with a small group of
assistants, a lab and a workshop. I did present a series of lectures
afterwards, twice in Europe. But it was all just in a day's work with
occasional trips aboard cable laying ships and at manufacturers. The nearest
I got to the States was Newfoundland and Nova Scotia. I then shifted in
succession to several entirely different fields of operations. But no
experience is ever lost.
--
Reg, G4FGQ