Richard Clark wrote:
You can still call it a transducer though - or a thigamajig.

Such a sophisticated concept deserves better. I suggest

"Triactuatedmulticomplicator" or TAMC for short.
--
73, Cecil http://www.qsl.net/w5dxp



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On Mon, 18 Aug 2003 14:20:49 -0500, W5DXP
wrote:

Richard Clark wrote:
You can still call it a transducer though - or a thigamajig.

Such a sophisticated concept deserves better. I suggest

"Triactuatedmulticomplicator" or TAMC for short.

Hi Cecil,

Given the high dudgeon that attends yet another inflammatory subject,
I would offer that antenna is enough - sheesh, didn't someone ask why
all the difficulty? It's not rocket surgery after all. :-)

73's
Richard Clark, KB7QHC

Richard Clark wrote in message . ..

Zc = 376.730 · ohms

That is, the Z of free space is expressed in exactly the same terms as
a carbon composition resistor. Now given the genesis of this debate
is that free space Z is somehow different from the expression of
Radiation Resistance (e.g. 37 Ohms for a monopole) the only possible
rhetorical objection is that free space is not lossy like a carbon
resistor (non-dissipative). Well, neither is the Radiation
Resistance! Even rhetoric fails. ;-)



Thank you Richard! Someone that's making sense on this NG after Roy lost
his sense!


An antenna is a structure that transforms Radiation Resistance into
the Impedance of free space, as shown, and by definition. Both use
identical MKS units, both are identical characteristics.


Ohms are still always Ohms, regardless of what you are measuring.
And it's very interesting that the E and H fields have units of
Volts/meter and Ampere(turn)/meter, which when you divide one by the
other, you get basically Volts/ampere, just like you would in a
transmission line.

But I don't claim that a wave traveling in a transmission line is
the same as a wave traveling through free space, even if Roy claims
this is what i mean.


Slick

Richard Clark wrote in message . ..

But this is repetition and evidence has been offered. As you have
revealed no new information that changes these relationships, nor have
you revealed any other representation of free space characteristic Z
in terms not already part of the MKS/SI Canon, then I am satisfied
that I will not change your mind.

73's
Richard Clark, KB7QHC


You may change Roy's mind, Richard, but he could never admit this
in public, because too many people are reading and it would make him
look bad.


Slick

"Dr. Slick" wrote:

Ohms are still always Ohms, regardless of what you are measuring.
And it's very interesting that the E and H fields have units of
Volts/meter and Ampere(turn)/meter, which when you divide one by the
other, you get basically Volts/ampere, just like you would in a
transmission line.

How do you know when the reduced units of one computation mean the
same thing as another?

An example:
The reduced units of modulus of elasticity (in/in/psi - psi) is
the same as the units for stress (psi) and yet modulus of elasticity
is clearly not stress. And in this case, the unreduced units are
much more descriptive than the reduced units. Reducing discards
information.

On the other hand, Torque (Newton*metres) when multiplied by
Radians (metre/metre) does give Energy (N*m*m/m - N*m), but only
after reduction. And for sure, Torque (N*m) is not the same as
Energy (N*m).

So sometimes it is appropriate to say the reduced results are the
same and some times it is not. Is there a way to know when it is
legal?

What rules have you used to conclude that reducing V/m/A/m to V/A
is appropriate?

....Keith

(Dr. Slick) wrote in message . com...

And your example of the product of RC being in seconds actually
makes sense! After all, it is an RC time constant! The number of
seconds it takes to get to 90% charge or discharge of a series RC
circuit.


Opps! Actually, for one RC time constant, it's about 36.79% of the
initial surge current when charging, or 63.21% of the final charged
voltage at t=infinity (it theoretically never reaches a full charge,
as you obviously know).

The point is that the units of R*C lead to units of seconds, which
is exactly what it should be. There is order to the universe!
Sometimes...


Slick

On Tue, 19 Aug 2003 06:07:19 -0400, wrote:

How do you know when the reduced units of one computation mean the
same thing as another?

They are ALWAYS fungible. You can certainly munge up operations to
prove otherwise, and it is easy to do with some really long chain of
computations.

I would suggest you investigate any of the several really good
Mathematics programs, one being Mathcad which offers a huge repository
of such Units tools that it uses to the enormous and enthusiastic
response by engineers and scientists. There greatest asset is in
allowing, you, the user, to enter your measurement in whatever Units
your profession is comfortable with, and marry them into a novel
situation at the interface to another discipline. How much horse
power generator is needed to supply electrical power that is required
to move a speaker cone how many inches to compress air to what sound
pressure level for it to be just barely discernable to the average
listener? That standard could be described as the force necessary to
move the eardrum the same distance as the diameter of an Hydrogen
Atom.


An example:
The reduced units of modulus of elasticity (in/in/psi - psi) is
the same as the units for stress (psi) and yet modulus of elasticity
is clearly not stress. And in this case, the unreduced units are
much more descriptive than the reduced units. Reducing discards
information.

The Elasticity Modulus is described in kg/mm² which is not quite MKS,
but performing the necessary operation to make it so does not remove
any information whatever. If we stick with electronics and discuss
the stress mechanics of piezos (crystals), then stress can be
described in terms of
(volts/meter) / (newtons/meter²)

On the other hand, Torque (Newton*metres) when multiplied by
Radians (metre/metre) does give Energy (N*m*m/m - N*m), but only
after reduction.

This is negative evidence? It is more a clouded argument.

And for sure, Torque (N*m) is not the same as
Energy (N*m).

You are confusing Work and Rotational Statics as being different.
Can you distinguish between Kinetic Energy and Potential Energy
described in mechanical units? If so, both are used to describe the
complete Work equation:
(KE2 - KE1) + (PE2 - PE1) = 0

So sometimes it is appropriate to say the reduced results are the
same and some times it is not. Is there a way to know when it is
legal?

Strictly speaking from the point of legality, it is demanded of
Professional Engineers by the National Institutes of Science and
Technology (what was called the National Bureau of Standards or NBS
years ago).

This means that ANY P.E. that describes a physical relation that does
not conform to these scientific concepts, and damage results to that
Professional Engineer's customer, then that P.E. is liable in a court
of law. This form of legality is the whole point of being P.E.s and
the government making the demand that P.E.s be part of describing
engineering codes and performing design review.

Anyone here who has put up a tower has had to jump through this hoop.
One of their principle concerns is found in the CM or Center of
Moment. The employment of Units transformation and reduction is part
and parcel to their activity (How tall? Length. How heavy? Mass and
the constant of Gravity).

The ONLY reason the city, or county insists on this report is to have
someone take the insurance hit if there is an error in meeting code.
Was your tower too tall for the weight of that long lever arm (your
antenna boom and elements) that also created a torque? Did it snap
with wind load? Did the guys snap through poor tensioning? Every one
of these uses Units that eventually boil down to one of the
tower/beam/guy specifications expressed in identical Units. There is
no other way for anyone to put their name to a report qualifying your
tower otherwise - why would they want to jack up their malpractice
payments? And it would be their insurance, not the city's if your
tower fell on a citizen and their heirs sued because of the city's
permission to you to erect it. This is called negligence and is why
your homeowner's insurance would walk away from you for a tower
collapse where the tower was not inspected to code.

If the P.E. met the standards of transforming between various systems
and observed the Physical Constants defined by NIST, then the P.E. is
NOT liable. If the P.E. is not liable, neither is the city/county.
If the P.E. and government are not liable, you have a problem.


What rules have you used to conclude that reducing V/m/A/m to V/A
is appropriate?

...Keith

Hi Keith,

You probably have no concern for the monolog about P.E.s or you put it
behind you long ago. Or so you and others might presume. That's fine
and this divergence off into mechanics may help some see the relations
but to answer your last question and keep it within the context of the
subject line, we should look at another reference that is less remote
than NIST and closer to antennas:
"Fields and Waves in Communication Electronics," Ramo, et al.

From page 3 (yes, pretty up front):

"Various systems of units have been used, but hat to be used in
this text is the International System (SI for the equivalent in
French) introduced by Giorgi in 1901. This is the
meter-kilogram-second (mks) system, but the great advantage
is that electric quantities are in the units actually measured:
coulombs, volts, amperes, etc."

This reference proceeds to describe those Physical Constants and their
relations that define Permittivity that I have already fully revealed
in a recent posting. If we were to proceed to page 71:

"The quantity known as the magnetic field vector or magnetic
field intensity is denoted H [sound familiar folks?- rwc] and
is related to the vector B define by the force law (2) through
a constant of the medium known as the permeability, µ:
B = µH
...
"In SI units, force is in newtons (N), Current is in amperes (A)
and magnetic flux density B is in tesla (T), which is weber per
meter squared or volt second per meter squared and is 10^4
times the common cgs unit, gauss. Magnetic field H
is in amperes per meter and µ is in henrys per meter. ...
The value for µ for free space is
µ0 = 4 · pi · 10^-7 · H/m"

So, there you have it. Absolutely identical to my other posts. The
RF engineering community's usage of free space Z is in full compliance
with the standards established and maintained by NIST. Both these
sources and standards are employed by commercial engineers and
Professional Engineers alike. It makes no sense to do otherwise
unless you want to start your own system of measurement that allows a
CFA to be 110% efficient. We get many efficiency claims that can ONLY
be judged through these associations described.

The chain of relationships proves that the "ohms" described by the Z
of free space are identical to the "ohms" used for ANY electrical
measurement, among which are the resistance determination of an
antenna (for any feed), or the resistance presented by a carbon
composition resistor.

73's
Richard Clark, KB7QHC

wrote in message ...
"Dr. Slick" wrote:

Ohms are still always Ohms, regardless of what you are measuring.
And it's very interesting that the E and H fields have units of
Volts/meter and Ampere(turn)/meter, which when you divide one by the
other, you get basically Volts/ampere, just like you would in a
transmission line.

How do you know when the reduced units of one computation mean the
same thing as another?

An example:
The reduced units of modulus of elasticity (in/in/psi - psi) is
the same as the units for stress (psi) and yet modulus of elasticity
is clearly not stress. And in this case, the unreduced units are
much more descriptive than the reduced units. Reducing discards
information.


Not really. Look at this:

http://hyperphysics.phy-astr.gsu.edu/hbase/permot3.html

If you notice, the strain is = delta L/ original L, so the strain
is dimensionless. So Young's modulus actually seems to represent the
N/m**2 (PSI) that is required to elongate something to twice it's
original length: delta L = original L, so that the denominator is 1.

interesting that you bring this up.



On the other hand, Torque (Newton*metres) when multiplied by
Radians (metre/metre) does give Energy (N*m*m/m - N*m), but only
after reduction. And for sure, Torque (N*m) is not the same as
Energy (N*m).



Hunh?? how did you get radians = m/m?

Look he

http://www.sinclair.net/~ddavis/170_ps10.html

I admit that this page reminded me that radians are
dimensionless.

So the torque times radians just gives you the work done, which
is in the same units as torque by itself. it's a bit confusing, but
Rotational units are used differently from linear ones (you have the
moment arm), so linear units are force is Newtons or lbs, and work is
in Newton*meters or ft*lbs.

I'm not totally sure, but the reason for this discrepancy seems
to be related to the fact that upon each rotation, you end up at the
same point, so in a certain sense, no work is done. But the crux is
that angles are dimensionless:

http://mathforum.org/library/drmath/view/54181.html

But in either case, rotational or cartesian, the Newton is still
a Newton, and so are the meters.




So sometimes it is appropriate to say the reduced results are the
same and some times it is not. Is there a way to know when it is
legal?

What rules have you used to conclude that reducing V/m/A/m to V/A
is appropriate?

...Keith


Basic algebra and cancellation of units. When have you found it
not to be appropriate? I'll admit that it can be a bit confusing
going from cartesian to rotational, and you have to understand the
context, but the UNITS ARE ALWAYS THE SAME. Isn't this the crux of
science and math? That we have certain standards of measurement, so
when we say it's a meter, it's a meter? God, i hope so.


Slick

"Dr. Slick" wrote:

wrote in message ...
"Dr. Slick" wrote:

Ohms are still always Ohms, regardless of what you are measuring.
And it's very interesting that the E and H fields have units of
Volts/meter and Ampere(turn)/meter, which when you divide one by the
other, you get basically Volts/ampere, just like you would in a
transmission line.

How do you know when the reduced units of one computation mean the
same thing as another?

An example:
The reduced units of modulus of elasticity (in/in/psi - psi) is
the same as the units for stress (psi) and yet modulus of elasticity
is clearly not stress. And in this case, the unreduced units are
much more descriptive than the reduced units. Reducing discards
information.


Not really. Look at this:

http://hyperphysics.phy-astr.gsu.edu/hbase/permot3.html

If you notice, the strain is = delta L/ original L, so the strain
is dimensionless.

Yes, and no. It was length per length, not, for example, volt per volt
or
pound per pound or ...

So dimensionless quantities are not all the same, even though they are
all dimensionless.

So Young's modulus actually seems to represent the
N/m**2 (PSI) that is required to elongate something to twice it's
original length: delta L = original L, so that the denominator is 1.

interesting that you bring this up.


On the other hand, Torque (Newton*metres) when multiplied by
Radians (metre/metre) does give Energy (N*m*m/m - N*m), but only
after reduction. And for sure, Torque (N*m) is not the same as
Energy (N*m).


Hunh?? how did you get radians = m/m?

Length of arc divided by radius in MKS units. How quickly we forget when
we get in the habit of leaving out all the units.

After multiplying Torque by Radians, you have computed the length
along the arc through which the force has acted - energy, of course.

Look he

http://www.sinclair.net/~ddavis/170_ps10.html

I admit that this page reminded me that radians are
dimensionless.

So the torque times radians just gives you the work done, which
is in the same units as torque by itself. it's a bit confusing, but
Rotational units are used differently from linear ones (you have the
moment arm), so linear units are force is Newtons or lbs, and work is
in Newton*meters or ft*lbs.

I'm not totally sure, but the reason for this discrepancy seems
to be related to the fact that upon each rotation, you end up at the
same point, so in a certain sense, no work is done.

Actually, you've done 2*pi*radius*force work. Moving one circumference
times the force.

But the crux is
that angles are dimensionless:

http://mathforum.org/library/drmath/view/54181.html

But in either case, rotational or cartesian, the Newton is still
a Newton, and so are the meters.


So sometimes it is appropriate to say the reduced results are the
same and some times it is not. Is there a way to know when it is
legal?

What rules have you used to conclude that reducing V/m/A/m to V/A
is appropriate?

...Keith

Basic algebra and cancellation of units. When have you found it
not to be appropriate?

It is not appropriate to consider Torque and Work to be the same, though
they have the same units.

It is not appropriate to consider modulus of elasticity and pressure
to be the same, though they have the same units after simplification.

But after multiplying Torque times Radians it is necessary to simplify
to discover that Work is the result.

I conclude that simplification is sometimes necessary and appropriate
but other times it is not. I am having difficulty knowing how to know
when it is appropriate.

This brings us back to the Ohms of free space and the Ohms of a
resistor.

While I don't know whether they are the same or not (and opinion seems
divided), it is clear that arguing that they are the same because the
units (after simplification) are the same is quite falacious. On the
other hand if the units were different, it would be clear that they
are not the same.

....Keith

Richard Clark wrote:
"So how do the "ohms" of free space differ from the "ohms" of a quarter
wave monopole`s radiation resistance?"

Terman says something like: the radiation resistance has a value that
accepts the same power as the antenna takes when the equivalent resistor
is placed in series with the antenna."

Roy Lewallen has already said that the resistance of free-space is the
ratio of the E-field to the H-field. Fields relate to the forces they
exert. No amps in empty space which has no electrons. Only when a
conductor is inserted is there a path for electrons to travel in.

Evidence that antenna impedance does not define radiation is the
identical radiation produced by antennas which are very different.

The folded monopole and the quarter-wave vertical are quite different.
The monopole is a small squashed loop. The quarter-wave vertical is a
single rod. Feed point resistance is 150 ohms for a typical folded
unipole and it is 28 ohms for the typical quarter-wave vertical. A look
at Arnold Bailey`s catalog shows identical radiation patterns and gain
for both antennas.

Best regards, Richard Harrison, KB5WZI