antenna length vs. dielectric constant?

[Guest]

Most antenna length formulas assume the antenna will be operating in either
air or a vacuum where the dielectric constant of "space" is equal to 1.

What happens to the length of a half-wave dipole when it is operated in a
space with a dielectric constant of 80?

Since the end-to-end capacitance increases with an increased dielectric constant
for the medium surrounding an antenna, I'm guessing that an antenna designed
for operation in air must be shortened to reduce both its incuctance and
end-to-end capacitance to regain the resonance it used to have in air. I
just don't know *how much* shorter.

Can somebody "do the math" for me?

Jim

 

[Don Pearce]

Most antenna length formulas assume the antenna will be operating in either
air or a vacuum where the dielectric constant of "space" is equal to 1.

What happens to the length of a half-wave dipole when it is operated in a
space with a dielectric constant of 80?

Since the end-to-end capacitance increases with an increased dielectric constant
for the medium surrounding an antenna, I'm guessing that an antenna designed
for operation in air must be shortened to reduce both its incuctance and
end-to-end capacitance to regain the resonance it used to have in air. I
just don't know *how much* shorter.

Can somebody "do the math" for me?

Jim

You just scale the whole thing in inverse ratio to the dielectric
constant; air = 1. Dielectric constant describes the relative velocity
of the wave in the medium. Seen this way, the scaling is clear.

But presumably at some point the wave must emerge from the Er80 medium
into Er1 of air. Have you given any thought to that transition, which
will be highly reflective?

d

Pearce Consulting
http://www.pearce.uk.com

 

[Larry Brasfield]

Most antenna length formulas assume the antenna will be operating in either
air or a vacuum where the dielectric constant of "space" is equal to 1.

What happens to the length of a half-wave dipole when it is operated in a
space with a dielectric constant of 80?

The velocity of wave propagation in space is predicted
by 1/sqrt(u * e) where 'u' and 'e' are the permability
and permittivity of free space. In your medium, the
wavelength will be reduced by a factor sqrt(80).

Since the end-to-end capacitance increases with an increased dielectric constant
for the medium surrounding an antenna, I'm guessing that an antenna designed
for operation in air must be shortened to reduce both its incuctance and
end-to-end capacitance to regain the resonance it used to have in air. I
just don't know *how much* shorter.

Just scale it according to the wavelength reduction.

Can somebody "do the math" for me?

It's simple enough that you can do it.

 

[Don Pearce]

The velocity of wave propagation in space is predicted
by 1/sqrt(u * e) where 'u' and 'e' are the permability
and permittivity of free space. In your medium, the
wavelength will be reduced by a factor sqrt(80).


Just scale it according to the wavelength reduction.


It's simple enough that you can do it.

Oops - forgot the square root in my post.

d

Pearce Consulting
http://www.pearce.uk.com

 

[James Meyer]

But presumably at some point the wave must emerge from the Er80 medium
into Er1 of air. Have you given any thought to that transition, which
will be highly reflective?

d

Pearce Consulting
http://www.pearce.uk.com

Actually, I'm working on a system to use a radio to link a submarine to
a transmitter. Both TX and RX antennas will be under water. Fresh water. I
know things get much more complicated in salt water.

Jim

 

[Reg Edwards]

Propagation velocity, in metres per second, along a thin wire is
proportional to 1/Sqrt(L*C) where L is the inductance and C is the
capacitance per metre.

Encasing the wire with a VERY LARGE, theoretically infinite, volume of a
dielectric material, including off the ends, will increase the capacitance
per metre by K, where K is the dielectric constant of the material.

The material will have NO effect on inductance.

Therefore, the velocity for a very large volume will be reduced by the
factor Sqrt(K).

But take the practical case of a wire encased with polyethylene only up to a
diameter of 25mm, 1" inch. Polyethylene has K = 2.6.

Because of the very small diameter of the dielectric relative to infinity,
the velocity along the wire will decrease by only a few percent. To maintain
the same resonant frequency, as on an antenna, the wire must be pruned to a
shorter length by the same percentage.

Calculations to find the exact percent reduction are complicated. The wire
diameter must also be taken into account.

Incidentally, for pure water, K=80. But who will want to encase a radio
antenna in many gallons of water just to reduce its length by a few more
percent. It is obvious that even a heavy tropical downpour will not have
the slightest effect. ;o)

 

[James Meyer]

Incidentally, for pure water, K=80. But who will want to encase a radio
antenna in many gallons of water just to reduce its length by a few more
percent.

Suppose you needed to communicate with a submarine and could put both
antennas underwater?

Jim

 

[Mac]

Suppose you needed to communicate with a submarine and could put both
antennas underwater?

Jim

You mean a full size submarine? In the ocean? Are you familiar with the
expression "lossy dielectric?"

Just wondering.

(yes I know they CAN communicate with subs, but they use low frequencies,
and the subs trail very long antennas to do it.)

--Mac

 

[Reg Edwards]

Most antenna length formulas assume the antenna will be operating in either
air or a vacuum where the dielectric constant of "space" is equal to 1.

What happens to the length of a half-wave dipole when it is operated in a
space with a dielectric constant of 80?

Since the end-to-end capacitance increases with an increased dielectric constant
for the medium surrounding an antenna, I'm guessing that an antenna designed
for operation in air must be shortened to reduce both its incuctance and
end-to-end capacitance to regain the resonance it used to have in air. I
just don't know *how much* shorter.

Can somebody "do the math" for me?

Jim

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

There is only a very remote probability that anyone beside myself who is
reading this can produce the math. Then there's the problem of writing the
formulae involved in HTML. You will be wasting your time hunting through
Google and books. And I havn't the many hours to spare to sort out your
problem and write out the solution. In any case, to avoid errors, it would
be necessary to write it all on paper with a pencil and post it to you via
snail mail in the hope you could understand it. You can see my predicament.
Sorry!

By the way, why do you wish to know?

 

[Don Pearce]

Actually, I'm working on a system to use a radio to link a submarine to
a transmitter. Both TX and RX antennas will be under water. Fresh water. I
know things get much more complicated in salt water.

Jim

No, salt water is really easy. It doesn't work, so you don't try. (Yes
I know about those ELF systems)

d

Pearce Consulting
http://www.pearce.uk.com

 

[James Meyer]

You mean a full size submarine? In the ocean? Are you familiar with the
expression "lossy dielectric?"

Just wondering.

--Mac

It's for a small model of a submarine that will be used to test a
prototype propulsion system and it will be deployed in a large fresh water
flooded stone quarry. A few hundred feet of range is all we need. That's all
I'm allowed say right now.

Jim

 

[Reg Edwards]

By the way, why do you wish to know?

===============================
If you had a capacitance formula what would you do with it?

Whatever radio frequency you have in mind for your dipole - forget it.

Try doing what whales do. Use a very low audio frequency, a loudspeaker and
a microphone. Or borrow a trained whale to read the meters. ;o)

 

[James Meyer]

There is only a very remote probability that anyone beside myself who is
reading this can produce the math.

Both your earlier post and the one from Larry Brasfield gave me a good
starting point. The rest I can extract empirically.

By the way, why do you wish to know?

Just what I've recently posted. If I went into more detail, we'd both
"dissappear".

Jim

 

[Rich Grise]

It's for a small model of a submarine that will be used to test a
prototype propulsion system and it will be deployed in a large fresh water
flooded stone quarry. A few hundred feet of range is all we need. That's all
I'm allowed say right now.

This sounds like a job for ultrasonic, if you can filter out bottom
clutter.

Good Luck!
Rich

 

[Mac]

It's for a small model of a submarine that will be used to test a
prototype propulsion system and it will be deployed in a large fresh water
flooded stone quarry. A few hundred feet of range is all we need. That's all
I'm allowed say right now.

Jim

Wow. Cool. It really does seem that ultrasonics would work better. But I
have never tried to transmit RF underwater, so I guess I should keep my
opinion to myself. ;-)

Anyway, as for your original question, it seems to me that,
theoretically at least, you can just use any typical land-based
antenna, but scale the dimensions by sqrt(80). So if you are planning
on using ~70 MHz, you could design a land antenna for ~630 MHz. I get
630 by pretending that sqrt(80) = 9. Then I multiply 70 MHz by 9 to
get 630 MHz.

If you are thinking of using frequencies in this range, maybe you could
just try a UHF TV antenna in the water.

An interesting question is what is the input impedance?

Best of luck to you.

--Mac

 

[Andy]

Andy replies,

Screw the math...... Get an SWR meter and go to a river....
The empirical evidence will tell you a lot more than an academic
exercise
will.

In this case, there is no substitute for field measurements....

If, after you find the answer, you can get it to agree with some
theoritical treatment, you have a good basis for an article, which
at $50 a page will pay you for your efforts....

Andy in Fink, Texas

 

[Reg Edwards]

Andy, you first have to get the equipments working before thinking about an
SWR meter, the most time-wasting instrument ever invented.

HF antennas submerged in sea water just don't work at all. In the absence
of anything else they may be very good as ground electrodes.

Stop trying to be clever and wasting the questioners's time. He already
knows more about it than you do - and even he doesn't know very much.

 

[Barry Lennox]

Both your earlier post and the one from Larry Brasfield gave me a good
starting point. The rest I can extract empirically.

Bear in mind that the model RC submarine crowd just use off-the-shelf
commercial RC transmitters to control model subs 2-300 hundred metres
away and to a few metres depth. Despite some saying it won't work, it
does. So maybe you won't have have to re-invent the wheel. Do a Google
search for model RC subs, there is a lot of info out there.

Barry Lennox.

 

[Mark]

read what the op said


he needs to transmit a few hundred feet

if he has to transmit only a few hundred feet, the antenna doesn't
have to "work very well"

even if it works poorly. it will work well enough

Mark

 

[James Meyer]

Bear in mind that the model RC submarine crowd just use off-the-shelf
commercial RC transmitters to control model subs 2-300 hundred metres
away and to a few metres depth. Despite some saying it won't work, it
does. So maybe you won't have have to re-invent the wheel. Do a Google
search for model RC subs, there is a lot of info out there.

Barry Lennox.

Thanks Barry. We've already used a standard RC setup to do the few
meters depth thing. With Larry and Reg's help, I have enough info to begin the
next stage of the project. I looked at the web pages for model subs almost a
year ago and their experiences matched our own at the time. Perhaps it's time
to take another look to see if there has been any progress since then.

Jim