Supporting theory that Antennas "Match" to 377 Ohms (Free space)
Dr. Slick
I looked through Albert Shadowitz's "The Electromagnetic Field",
and found on page 554 support for my original statement that antennas
"match" to the impedance of free space (377 Ohms).
I didn't buy the book, but professor Shadowitz did write on this
page about how creating antennas to most efficiently transfer power to
free space is a similar problem to matching a circuits source to its
load. He goes on to make a short comparison between source/load
impedances to an antenna matching to the impedance of free-space.
Food for thought and no doubt, debate.
Slick
WB3FUP (Mike Hall)
is actually science to support use of a 300 ohm antenna. who'd a thunk it.
--
73 es cul
wb3fup
a Salty Bear
"Dr. Slick" <radi...@aol.com> wrote in message
news:1d15af91.0308...@posting.google.com...
Yuri Blanarovich
Yep,
I tried to question that few moons back, but was "convinced" here that it is
not important.
Generally antenna exhibits all kinds of impedances along its length. I was
reasoning that antenna having its lowest impedance higher or closer to 377 ohm
should have better efficiency in coupling to the space (air). So loops and
folded dipoles should be better in that respect. K8CFU et al, when doing
experiments with verticals and radials, found that folded monopole measured
higher signal levels (over simple monopole) than expected.
Any progress since then?
Yuri, K3BU
Richard Harrison
"---creating antennas to most efficiently transfer power to free space
is a similar problem to matching a circuit`s source to its load. (Quote
from Shadowitz)"
Reciprocity rules in antennas. Kraus has an Apendix D (Absorbing
Materials) to his "Antennas For All Applications".
Kraus says:
"The use of space cloth (Z=377 ohms per square) placed lambda/4 from a
reflecting plane was invented by Winfield Salisbury (1) at Harvard Radio
Research Laboratory during WW-2 ---." (Shades of stealth)
However, a century of antenna experimentation has not revealed a
practical need to especially design an antenna to match its radiation to
a resistive 377 ohms.
Most effective radiation occurs when the antenna circuit is matched and
has no loss. 100% of the energy accepted is radiated.
Best regards, Richard Harrison, KB5WZI
Reg Edwards
impedance of its feedline, or whether it's matched to it or not.
Now I suppose somebody will drag in the irrelevant matter of SWR on the
feedline.
---
Reg, G4FGQ
W5DXP
> Most effective radiation occurs when the antenna circuit is matched and
> has no loss. 100% of the energy accepted is radiated.
In the real world for the same size wire on HF, a folded dipole should
be slightly more efficient than a dipole because of lower I^2*R losses.
--
73, Cecil, W5DXP
W5DXP
Actually, you brought up the subject. :-) The feedline's power transfer
efficiency is just as important as the antenna's radiating efficiency.
--
73, Cecil, W5DXP
Dick Carroll;
Reg Edwards wrote:
The reason they might do that would be concern for the antenna *system*
efficiency and not
just that of the antenna.
Dick
Reg Edwards
> > feedline.
>
> Actually, you brought up the subject. :-) The feedline's power transfer
> efficiency is just as important as the antenna's radiating efficiency.
> --
Cecil, you forgot the efficiency of the PA DC power supply. There's much
more energy wasted there than what's lost in the feeder.
Sort out your power budget. ;o)
---
Reg.
Richard Clark
<wb3...@msn.com> wrote:
>And another good reason the folded dipole FM antenna works so great. There
>is actually science to support use of a 300 ohm antenna. who'd a thunk it.
Hi Mike,
What you describe is the feedpoint Z not the antenna Z which to all
intents and purposes is not far from the original, single-wire dipole.
73's
Richard Clark, KB7QHC
W5DXP
>>Actually, you brought up the subject. :-) The feedline's power transfer
>>efficiency is just as important as the antenna's radiating efficiency.
> Cecil, you forgot the efficiency of the PA DC power supply. There's much
> more energy wasted there than what's lost in the feeder.
My DC power supply is a 12 VDC marine battery charged by a solar
panel. What's the efficiency of free energy? :-)
> Sort out your power budget. ;o)
Actually, I don't much care about the efficiency of the electronics.
60 Hz energy is cheap. I am much more interested in getting the
generated RF into the Æther.
--
73, Cecil, W5DXP
Reg Edwards
> generated RF into the Æther.
>
Cec, Good, now we're back on track. But what's the 377 ohms of nothingness
to do with a random length of wire which has any impedance you fancy just by
connecting to it. Even without making a connection and just using your
imagination. Could the person who confidently raised this subject from the
dead please give us some clues about calculating the turns ratio. I don't
have ready access to the works of Maxwell and there's no mention of it in my
1992 edition of the ARRL Handbook.
And I think my smiley is better than yours!
Reg Edwards
>What you describe is the feedpoint Z not the antenna Z which to all
>intents and purposes is not far from the original, single-wire dipole.
>
>
After all these years you're catching on. What you really meant to say was
that the feedpoint impedance is not the same thing as the radiation
resistance.
'Antenna' impedance' in the present context is not a phrase known to radio
engineering. Please define.
---
Reg
Dave Platt
Reg Edwards <g4fgq...@btinternet.com> wrote:
>Cec, Good, now we're back on track. But what's the 377 ohms of nothingness
>to do with a random length of wire which has any impedance you fancy just by
>connecting to it. Even without making a connection and just using your
>imagination. Could the person who confidently raised this subject from the
>dead please give us some clues about calculating the turns ratio.
I think you have to measure the diameter of free space first, before
you can calculate the turns ratio.
Do write, when you get to the far side, and let us know how the
weather is, OK? ;-)
--
Dave Platt <dpl...@radagast.org> AE6EO
Hosting the Jade Warrior home page: http://www.radagast.org/jade-warrior
I do _not_ wish to receive unsolicited commercial email, and I will
boycott any company which has the gall to send me such ads!
Richard Clark
wrote:
>
>Yep,
>I tried to question that few moons back, but was "convinced" here that it is
>not important.
It is only an academic exercise as long as you use wire.
>Generally antenna exhibits all kinds of impedances along its length.
Very true, but that is simply an observation of fact that has nothing
to do with purpose.
>I was
>reasoning that antenna having its lowest impedance higher or closer to 377 ohm
>should have better efficiency in coupling to the space (air).
Just by virtue of using a conductor that can support its weight in
suspension (or compression if a vertical); this generally guarantees
little Ohmic loss which is the only source (other than ground) of
degrading efficiency.
> So loops and
>folded dipoles should be better in that respect. K8CFU et al, when doing
>experiments with verticals and radials, found that folded monopole measured
>higher signal levels (over simple monopole) than expected.
>Any progress since then?
>
>Yuri, K3BU
Hi Yuri,
Like Mike, you are confusing feedpoint Z with radiation Z. For simple
designs like the half wave dipole and the quarter wave monopole, the
two (feedpoint Z and radiation Z) are nearly identical. Fold these
with extra wire, and all you are doing is Z transforming that value to
another value and doing nothing about the inherent loss (which is
trivial in the first place if your antenna is built from a conductor).
The forced debate that an antenna is/is-not an Impedance transformer
seems myopic in the extreme when simple requirements are violated.
Namely, you cannot offer a simple description of any component whose
dimension exceeds a tenth wavelength (and even that is too long for
these "debates"). It is absurd to think of the antenna somehow ending
at the tips of the wire as though alligator clips there would normally
be expected to connect it to the æther. This is like thinking a
speaker emanates sound only from the fringe ring of the cone. Neither
example could be more absurd (but I've no doubt there will be many
attempts to shatter that illusion).
An antenna is a transformer (or transducer for those who prefer) which
is immersed in a medium, not connected to it at the ends of the wire.
There is a continuum of connectivity that is moderated by complex
voltage and current relationships that are in turn intertwined with
complex field relationships (all being irrevocably related). This
complexity does not negate the transformer (or transduction)
operation. In fact the statement of the complexity is the answer to
the transform.
An antenna is a leaky transmission line. That the spreading of a
transmission line transforms its Z is a commonplace of long standing.
That a principle requirement be that this transition occur over a long
interval (again, the nature of wavelength is central) is also long
understood. Combine the spreading, over a long distance, with
deliberate intent to leak the "confined" fields is also a principle
found in practice. A notable example is found in the V antenna
described in other threads, and the venerable Rhombic is another.
Now, to retract those provisos of long distance (wavelength) and
simply look at the standard dipole or monopole; it is obvious that
they are violated, and yet this is only to the notion of a simple
transform. The short transition, leaky, transmission lines simply
have a more complex transform - not the absence of a transform. The
first obvious evidence is the notable STANDING WAVE found on the short
elements in glaring contrast to the bigger designs.
For some reason, many want to dismiss the arguments against standing
waves on transmission lines (the circulating currents are lost to the
Ohmic resistance of the line) as not being applicable to the radiating
(leaky) elements. Such magic. The simple fact of the matter is that
these radiating elements are often many, many times larger that the
leads that feed them and offer very much less loss for the exhibited
SWR. The path to dismissing the concern is less magic than the simple
practicality of having more copper.
A visual correlative of these points are illustrated at:
http://home.comcast.net/~kb7qhc/antennas/Fields/index.htm
Where I show not the far field characteristics, but the near field
interaction where the transformer operation takes place. The visual
element portrays the near fields in terms of Z in relation to that of
free space Z (regardless of what "Ohms" is thought to mean for
whatever context).
A point in example with a vertical
http://home.comcast.net/~kb7qhc/antennas/Fields/Vertical/index.htm
reveals that fields in the proximity about half way up the radiator
are nearly matched to the Z of free space. The complex transformer
has established a locality of no reflection. Likewise, at the base of
the same monopole, the local fields are in high turmoil. At a remote
distance where the far field characteristics are felt, these two
characteristics are washed out to present the standard radiation
pattern.
So what does it mean to have high values or low values? If we were to
consider the example of the dipole
http://home.comcast.net/~kb7qhc/antennas/Fields/Dipole/index.htm
We would see the monopole turned on its side, and mirrored with the
twin radiator. And then we would notice more, like ground (which was
part of the monopole but invisibly balanced). One should notice that
ground exhibits high SWR regions in its vicinity directly below the
feed point of the radiator. A bit of slight of hand here, because
ground is not magic either; and is simply another, albeit lossier,
radiator. At this wavelength spacing, the fact that there is high SWR
in the nearby fields, and that ground is reflective should come at no
great surprise.
To extend this observation we need only lift that dipole to half wave
to see if ground follows in its wavelength relationships as radiator.
It does. The near fields close to ground are obviously different in
nature, but what does this have to do with high or low values? Let's
slip a conductive ground mat below the dipole. The result is
illustrated as a radiator in a virtually calm space of field
interaction (and efficiency measurements show a ground screen improves
this quality).
The variations of physical structure all obviously impact near fields
and present every bit of evidence to the action of transformation (or
transduction). There are fields that create localities of low SWR or
high SWR and these are immaterial to the argument of reducing the
description of an antenna to transformer (or transducer) because the
nature of size imposes no "point" which answers where the transform
(or transduction) occurs except at the obvious feedpoint.
But I can well imagine there are those who still linger with the
question "what significance is there in these Hi/Low SWR values?"
Think back to your experience of driving a hot summer road, where the
shimmer ahead of you presented the vision of a blank blue in place of
the black asphalt. A common experience of the mirage, or temperature
inversion for those less inclined to the mystical explanation. It is
tempting to offer this media dislocality to the analogy of lensing the
radiation. For an antenna, this would be to the neglect of the same
requirement that the lens be much larger than the source however and
such is not the case with these smaller local effects. Instead, they
could be argued to be atropic aberrations (like a cloudy distorted
lens or sheet of glass, to extend the metaphor). As such, the ground
mat offers a means to remove some of the clouding (and is demonstrated
by an overall increase in radiation - better efficiency).
Bob
Couldnt resist
Rather than arguing about it why dont all of you construct many kinds of
dipoles and folded dipoles, then measure the radiated field strength. You
could then collate the info between you all and present it to the NG
<grin>
Bob VK2YQA
W5DXP
> Cec, Good, now we're back on track. But what's the 377 ohms of nothingness
> to do with a random length of wire which has any impedance you fancy just by
> connecting to it.
The ratio of the radiated E-field to H-field has no other choice.
If you stuff EM radiation into free space, the ratio of E-field to
H-field is 376.7 ohms. Zero energy is lost from the EM spectrum when
an electron throws off a photon (until that photon is annihilated).
--
73, Cecil http://www.qsl.net/w5dxp
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Roy Lewallen
of nearly any value (magnitude and phase).
Roy Lewallen, W7EL
William E. Sabin
The near field has a reactive component to the
impedance. But is it true that the real part of
that complex impedance must be 376.7 ohms resistive?
Bill W0IYH
Roy Lewallen
Not at all. For example, the magnitude of of the wave impedance E/H is
much lower than 377 ohms very close to a small loop, and much higher
than 377 ohms very close to a short dipole. Interestingly, as you move
away from a small loop, the magnitude of E/H actually increases to a
value greater than 377 ohms, then slowly approaches 377 ohms from the
high side as you move even farther away. The opposite happens for a
short dipole -- the E/H ratio drops below 377 ohms some distance away (a
fraction of a wavelength), then increases to 377 ohms as you go farther yet.
Roy Lewallen, W7EL
William E. Sabin
>
>>
>> The near field has a reactive component to the impedance. But is it
>> true that the real part of that complex impedance must be 376.7 ohms
>> resistive?
>>
>> Bill W0IYH
>
>
> Not at all. For example, the magnitude of of the wave impedance E/H is
> much lower than 377 ohms very close to a small loop, and much higher
> than 377 ohms very close to a short dipole. Interestingly, as you move
> away from a small loop, the magnitude of E/H actually increases to a
> value greater than 377 ohms, then slowly approaches 377 ohms from the
> high side as you move even farther away. The opposite happens for a
> short dipole -- the E/H ratio drops below 377 ohms some distance away (a
> fraction of a wavelength), then increases to 377 ohms as you go farther
> yet.
>
> Roy Lewallen, W7EL
>
There seems to more explanation needed.
If a lossless dipole is loaded with 100 W of
*real* power, that is the real power in the far
field, and it is also the real power very close to
the antenna, regardless of the type of antenna.
The value of real power is the same everywhere.
Since real power is in the real part of the
impedance, then how does the value of real
impedance (not the magnitude of impedance) vary
with distance from the antenna?
It seems that very close to (but slightly removed
from) the antenna the real part of the resistive
space impedance is nearly the same as the real
part of the driving point impedance of the
antenna. This real part is then transformed to 377
ohms (real) within the near field, suggesting that
the open space adjacent to the antenna performs an
impedance transformation. The near-field reactive
fields perform this function in some manner.
The real power radiated is found by integrating
the Poynting vector slightly outside the surface
of the antenna, and is equal to the real power
into the (lossless) antenna. This value is
constant everywhere beyond the antenna.
Bill W0IYH
Richard Clark
<sabinw@mwci-news> wrote:
>It seems that very close to (but slightly removed
>from) the antenna the real part of the resistive
>space impedance is nearly the same as the real
>part of the driving point impedance of the
>antenna. This real part is then transformed to 377
>ohms (real) within the near field, suggesting that
>the open space adjacent to the antenna performs an
>impedance transformation. The near-field reactive
>fields perform this function in some manner.
>
The figures at:
http://home.comcast.net/~kb7qhc/antennas/Fields/Dipole/index.htm
illustrate just how the dipole's near-field reactance maps out
(without respect for phase, and expressed in SWR relative to free
space Z).
Note that employing the term transform and antenna within the same
context is not de rigueur. ;-)
William E. Sabin
>
>
> Note that employing the term transform and antenna within the same
> context is not de rigueur.
It is 100 percent correct and appropriate.
Bill W0IYH
Roy Lewallen
the impedance", and in the last, that it's "found by integrating the
Poynting vector slightly outside the surface of the antenna". The
impedance is E/H, the Poynting vector E X H. Clearly these aren't
equivalent.
The radiated power is, as you say, the integral of the Poynting vector
over a surface. (And the average, or "real", radiated power is the
average of this.) The integral doesn't need to be taken slightly outside
the surface of the antenna, but can be any closed surface enclosing the
antenna. There's no necessity for E/H, or the real part of E/H, to be
constant in order to have the integral of E X H be constant.
The driving point impedance of the antenna depends on where you drive
it, and it bears no relationship I know of to the wave impedance (which
is, I assume, what you mean by "resistive space impedance") close to the
antenna. If you find any published, modeled, measured, or calculated
support for that contention, I'd be very interested in it.
Roy Lewallen, W7EL
David Robbins
impedance and also the performance of the antenna. some possible examples
that show this effect are the changes in element sizes when modeling an
antenna printed on a dielectric circuit board material or sandwiched in a
dielectric media. the change in wire length due to insulation is another
example, the dielectric properties of the insulation change the E/H ratio
near the wire. some examples may be found in many electromagnetics texts,
look at things like dielectric waveguides, or dielectrics in waveguides,
wires in dielectric media. even the detailed calculation of fields within a
dielectric filled coaxial cable should show this effect, change the
dielectric and you change the characteristic impedance... a measurable
effect from changing the 'space impedence' between the wires.
"Roy Lewallen" <w7...@eznec.com> wrote in message
news:vjdantr...@corp.supernews.com...
Yuri Blanarovich
>
>The driving point impedance of the antenna depends on where you drive
>it, and it bears no relationship I know of to the wave impedance (which
>is, I assume, what you mean by "resistive space impedance") close to the
>antenna.
We can look at the lowest impedance in particular antenna, which will have
higher impedance points elsewhere along its length. Looking at different
antennas or arrays we can have antennas with higher lowest impedance. Like
folded dipoles and loops. Would that not indicate and provide closer "match" to
free space impedance? Again, K8CFU measured that folded monopole "surprisingly"
gave higher field strengths than simple monopole radiator. Wouldn't that
indicate that there is something "wrong" (good) about higher impedance
antennas? Capture area reflected in here?
Yuri, K3BU
Roy Lewallen
the flow of the Earth's Chi.
Roy Lewallen, W7EL
Richard Clark
I happen to agree, but being correct and appropriate has nothing to do
with de rigueur. The subject line, after all, has long been abandoned
to the whim of fashion.
Dr. Slick
> actually i would expect that a change in E/H would change the driving point
> impedance and also the performance of the antenna. some possible examples
> that show this effect are the changes in element sizes when modeling an
> antenna printed on a dielectric circuit board material or sandwiched in a
> dielectric media. the change in wire length due to insulation is another
> example, the dielectric properties of the insulation change the E/H ratio
> near the wire. some examples may be found in many electromagnetics texts,
> look at things like dielectric waveguides, or dielectrics in waveguides,
> wires in dielectric media. even the detailed calculation of fields within a
> dielectric filled coaxial cable should show this effect, change the
> dielectric and you change the characteristic impedance... a measurable
> effect from changing the 'space impedence' between the wires.
>
Agreed, and emmersing a waterproof antenna into water will also
affect the input impedance.
Slick
William E. Sabin
> In the fourth paragraph, you say that "real power is in the real part of
> the impedance", and in the last, that it's "found by integrating the
> Poynting vector slightly outside the surface of the antenna". The
> impedance is E/H, the Poynting vector E X H. Clearly these aren't
> equivalent.
>
> The radiated power is, as you say, the integral of the Poynting vector
> over a surface. (And the average, or "real", radiated power is the
> average of this.)
Correction "real part of Poynting vector" noted.
The problem remains:
How is the *real* part of the antenna input
impedance, regardless of how it is fed and
regardless of what kind of antenna it is, get
"transformed" to the *real* 377 ohms of free space?
I believe (intuitively) that the reactive E and H
near-fields collaborate to create an impedance
transformation function, in much the same way as a
lumped-element reactive L and C network. In other
words, energy shuffling between inductive and
capacitive fields do the job and the E and H
fields modify to the real values of free space.
The details of this are murky, But I believe the
basic idea is correct.
Bill W0IYH
W5DXP
> I believe (intuitively) that the reactive E and H near-fields
> collaborate to create an impedance transformation function, in much the
> same way as a lumped-element reactive L and C network. In other words,
> energy shuffling between inductive and capacitive fields do the job and
> the E and H fields modify to the real values of free space. The details
> of this are murky, But I believe the basic idea is correct.
_Optics_, by Hecht, has a section 2.10 - Cylindrical Waves.
There is an interesting statement in that section: "No solutions
in terms of arbitrary functions can now be found as there were
for both spherical and plane waves."
The net reactive impedance component on a standing-wave antenna
is the result of the superposition of forward and reflected waves
on the standing-wave antenna. Presumably, a traveling-wave antenna,
like a terminated Rhombic, doesn't have reactive impedance components.
So my question is: Since the voltage and current are always in phase
in a traveling-wave antenna, is the near field of a traveling-wave
antenna ever reactive?
--
73, Cecil, W5DXP
Reg Edwards
> in a traveling-wave antenna, is the near field of a traveling-wave
> antenna ever reactive?
Cec, you're leading yourself astray again. What's reactance to do with
anything other feedpoint impedance?
Stand at a distance from a very long Beverage antenna. Focus your attention
on a particular half-wave length of it.
The voltage at one end of the half-wave will whizz up and down at a
frequency of x megahertz.
At the the other end of the half-wave length the voltage will whizz down and
up at x megahertz, ie., in time-antiphase with it.
Therefore, from where you are standing, the half-wavelength of wire will
behave and radiate exactly like a half-wave dipole. You have no means of
knowing whether there are standing waves along the wire or not. And clearly
it doesn't matter. To segregate antennas between standing-wave and
non-standing-wave types can be misleading.
To continue with the Beverage. Adjacent 1/2-wavelengths of wire form a
co-linear array are in antiphase with each other. Therefore there is no
broadside radiation from a long Beverage which contains an even number
number of halfwavelengths. There is a sharp null at an angle of 90 degrees
from the wire and as overall length increases so does the number of lobes in
the general direction of the wire.
This is just the opposite of a co-linear array, a standing-wave antenna,
along which the successive half-wave dipoles are all in time-phase with each
other.
But both types of antenna incorporate radiating 1/2-wave dipoles. And if
the near-field of one type has a reactive near-field (whatever THAT means)
then so must the other.
If there are no standing waves it does NOT mean the voltage along the whole
length of line or antenna is whizzing up and down in simultaneous time-phase
in which case there would indeed be a non-reactive near field. But neither
could there be any length or time delay involved.
Don't confuse instantaneous RF volts over a cycle with the envelope which
may remain constant or vary with time or distance.
----
Reg, G4FGQ.
Richard Clark
<sabinw@mwci-news> wrote:
>The problem remains:
>
>How is the *real* part of the antenna input
>impedance, regardless of how it is fed and
>regardless of what kind of antenna it is, get
>"transformed" to the *real* 377 ohms of free space?
>
Hi Bill,
Transformation, as a term, seems to be problematic without any more
care for the preferred term of transduction (ignoring the historical
usage it clashes with). How words could have any bearing on the
process itself is more a calmative to the user than a need for the
group.
So, if we were to simply ignore ALL the terms, how many show up at the
table to discuss the PROCESS (I hope that's the right word...)?
If we simply cast off the electrical aspect of it (seeing how
difficult it is to conduct discussion for this topic in that
vernacular), the correlative of the organ pipe would be useful. It
too creates a standing wave at the drive point; and it employs a
resonant structure wherein the wave stands. It conforms to the
transmission line principles of termination in that a close or open at
the end is meaningful, and harmonically related to wavelength in a
media. If this seems an outrage (because the former kidnapping of
terms is ignored) consider the following quote from Reference Data for
Radio Engineers:
"...Maxwell's initial work on electrical networks
was based on the previous work of Lagrange
in dynamic systems."
This reference then tumbles into the discussion of "Acoustic and
Mechanical Networks and their Electrical Analogs"
It can be seen that the structure imposes critical significance in the
harmonic component, but is wholly inert without excitation. In other
words, it is not the causative agent, nor is it the agent of
transmission. The pressure excess would cause air flow with or
without it. Of course, there is an efficiency problem in that lax
attitude and that necessarily brings us back to structure and fields
(pressure in this case).
What has this to do with near field and far field? For the organ
pipe, what is the near field, what is the far field? Here, we get
into issues as we formerly did by looking at dimension and wavelength.
There are two classes of Acoustic Impedance that bear to this
intimately.
Those two classes compute for a spherical wave front, and for a planar
wave front:
"...the acoustic impedance for a spherical
wave has an equivalent electrical circuit
comprising a resistance shunted by an inductance.
In this form, it is obvious that a small spherical
source (r is small) cannot radiate efficiently since
the radiation resistance [formula] is shunted by a small
inductance [formula]."
The plane wave Acoustic Impedance formula does not exhibit this
inductive shunt. The difference between the two cases is simply a
matter of scale, and is as arbitrarily chosen as with the abandoned
antenna. That is to say, the definition of antenna far field being
expressed as residing 10 wavelengths away also finds the correlative
in this difference of Acoustic Impedance.
What is this shunt? The compressibility of the medium which is the
mechanical analog of storage.
What is the difference between the case of the organ pipe and the
antenna? For the pipe, the medium is lossy (and employing a vacuum
brings its own obvious issues for the organ) and we find the loss
expressed in phonons (the heat of jostling material). For the antenna
(especially in the void of a vacuum, a useful medium) we find no such
issue and consequently no related phonons (loss to heat within the
medium). Some would note this also encompasses the traditional
demarcation between transducer and transformer.
Irrespective of the difference, both exhibit a region wherein the
MEDIA supports the transition (and perhaps we should call these
structures transitioners --- only kidding :-).
As I stated in the past, it is absurd to crop the picture such that
the description demands that an antenna ends at the literal tips of
its structure as if virtual clips connect it to the æther.
W5DXP
>>So my question is: Since the voltage and current are always in phase
>>in a traveling-wave antenna, is the near field of a traveling-wave
>>antenna ever reactive?
>
> anything other feedpoint impedance?
How can I lead myself astray by asking a question? :-)
Forget reactance. The question remains: Can the E-field and H-field
be out of phase if the voltage and current are in phase?
--
73, Cecil, W5DXP
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Roy Lewallen
field at any point in space you'd like. EZNEC reports phase angles of
the E and H fields, so it won't take long for you to find out.
Roy Lewallen, W7EL
William E. Sabin
> Roy Lewallen wrote:
>
>> In the fourth paragraph, you say that "real power is in the real part
>> of the impedance", and in the last, that it's "found by integrating
>> the Poynting vector slightly outside the surface of the antenna". The
>> impedance is E/H, the Poynting vector E X H. Clearly these aren't
>> equivalent.
>>
>> The radiated power is, as you say, the integral of the Poynting vector
>> over a surface. (And the average, or "real", radiated power is the
>> average of this.)
>
>
> Correction "real part of Poynting vector" noted.
>
> The problem remains:
>
> How is the *real* part of the antenna input impedance, regardless of how
> it is fed and regardless of what kind of antenna it is, get
> "transformed" to the *real* 377 ohms of free space?
>
> I believe (intuitively) that the reactive E and H near-fields
> collaborate to create an impedance transformation function, in much the
> same way as a lumped-element reactive L and C network. In other words,
> energy shuffling between inductive and capacitive fields do the job and
> the E and H fields modify to the real values of free space. The details
> of this are murky, But I believe the basic idea is correct.
>
> Bill W0IYH
>
For example, consider an EZNEC solution to an
antenna, say a 50 ohm dipole. The far-field 377
ohm solution provided by the program is precisely
the field that I am thinking about. How does
EZNEC, with its finite-element, method-of-moments
algorithm, transform a 50 ohm dipole input
resistance to 377 ohms in free space?
I don't want the equations, I want a word
description (preferably simple) of how EZNEC
performs this magic.
The far-field E and H fields are different from
the near-field E and H fields. What is going on?
Bill W0IYH
Richard Clark
<sabinw@mwci-news> wrote:
>
>The far-field E and H fields are different from
>the near-field E and H fields. What is going on?
>
Hi Bill,
The continuum of the structure presents a delay (by "moments" to use
the vernacular of MOM) that combines with all "moments" of the
previously existing and "near" separated field(s) to cause local
free-space media fluctuations in Z. At a greater distance, such
differences become trivial.
The local fields present a non-homogenous free-space media, some of
which is transparent, some of which is reflective, much of it
somewhere in between. The antenna distorts the medium it resides in
presenting much the same effect as gravity distorting the space-time
continuum. This is a leap of faith, certainly, but offers a
visualization that may be familiar. In optics it would be something
like dispersion where the structure is smaller than the wavelength
exciting it.
Roy Lewallen
The following description is a very simplified version of how NEC works.
I believe the whole NEC-2 manual is available on the web, for anyone who
wants a deeper and surely more accurate explanation.
First, an impedance is calculated for each segment of each wire, and a
mutual impedance for every segment relative to every other segment. This
is done in a rather complex way by assuming that each segment has sine,
cosine, and constant currents, calculating the field from each segment
arriving at each other segment, and evaluating the current induced on
the other segment by it. These impedances are put into a matrix, then
the currents on each segment are found by solving Ohm's law in matrix
form, where the E is provided by the specified sources. Once the
currents are found, the impedance at each of the sources is known. The
field from each segment is computed from the known current and assumed
current distribution along the segment with an approximate integral
equation that's solved numerically. The impedance of the medium (fixed
at free space in NEC-2 but user selectable in NEC-4) is of course
involved in this calculation, as it is for the mutual impedance calculation.
The fields are summed to obtain the overall field (both E and H) at any
point the user specifies. Both are reported in a near field analysis
output. In a far field calculation, the distance of the observation
point to all segments is assumed to be the same, and only the E field is
calculated.
An excellent and easy to follow description of the method of moments can
be found in Kraus' _Antennas_, Second Ed. I assume it's in the third
edition also, but it's not in the first. The NEC-2 manual recommends
R.F. Harrington, _Field Computation by Moment Methods_ (McMillan, 1968)
but I haven't seen this book.
I've tried to point out on this thread that although the feedpoint
impedance is an impedance with the units of ohms, and the impedance of a
plane wave in free space also has the units of ohms, they're not the
same thing. Feedpoint impedance is the ratio of a current to a voltage.
Wave impedance, or the intrinsic impedance of a medium, is the ratio of
an E field to an H field -- it's also the square root of the ratio of
the medium's permeability to its permittivity. An antenna converts
currents and voltages to E and H fields, it doesn't just transform one
impedance to another. Hence my insistence on calling an antenna a
transducer rather than a transformer.
Any explanation of an antenna as a transformer will have to include
parasitic array elements, which have zero feedpoint impedance, and array
elements that have negative feepoint resistances.
The answer to your last question is beyond my ability to answer. It's
discussed in great detail in most electromagnetics and antenna texts.
Roy Lewallen, W7EL
William E. Sabin wrote:
Reg Edwards
You had led yourself astray long before you asked the question or you
wouldn't have asked it. ;o)
-----------------------------------------------------------------
> Forget reactance.
-----------------------------------------------------------------
Good! I can't see why you dragged the reactance of fields in space into to
the arena.
-----------------------------------------------------------------
>The question remains: Can the E-field and H-field
> be out of phase if the voltage and current are in phase?
I answered your original question by saying that the near field due to a
halfwavelength of wire along an antenna with standing waves (like a
co-linear array of in-phase dipoles) was just the same as a halfwavelength
along a wire in an antenna without any standing waves (like a Beverage).
What applies to one must also apply to the other.
You have forgotten that fields exist in 3-dimensional space as well as in
time. Movement (velocity) is also directly involved. Phase relationships
between the various vectors depends on the direction (another vector) from
which the observer (that's you) makes his observations.
Whereas, when you introduce volts and current in phase, you are referring
only to a single point (in a straight length of wire) which has NO
dimensions. It's just a location in space somewhere along a transmission
line which is what an antenna wire actually is.
From occasional random sampling of the heated arguments presently taking
place on this newsgroup it may be deduced that most of you baffle-gabbers
don't have foggiest notions of what you are waffling about and there's no
hope of you ever doing so unless you go back to school and the 12-times
table and continue till you are familiar with Oliver Heaviside's
mathematical description of electromagnetic radiation. There are only 3
simultaneous differential equations to contend with. As the ancient Iraqis
and Greeks understood, all is a matter of Geometry in Motion. And THEY
didn't have the benefit of Eznec.
I notice Roy has abandoned you to the wolves (CB-ers and M3-ers). His
patience and hard labour have been exemplary.
Has that 50 million of you all got your electic lights fixed yet ? ;o)
Cecil excuses himself by saying it's the Devil who makes him do it. In my
case it's only 3/4 of a bottle of Bulgarian red plonk.
----
Reg, G4FGQ, Hic!
>
W5DXP
> I answered your original question by saying that the near field due to a
> halfwavelength of wire along an antenna with standing waves (like a
> co-linear array of in-phase dipoles) was just the same as a halfwavelength
> along a wire in an antenna without any standing waves (like a Beverage).
Unfortunately, that cannot be true. The maximum H-field amplitude at a current
null cannot possibly be the same as the maximum H-field where a current null
never exists.
--
73, Cecil http://www.qsl.net/w5dxp
-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
Reg Edwards
current
> null cannot possibly be the same as the maximum H-field where a current
null
> never exists.
Dear Cec, your nitpicking is inexhaustible and at this time in the morning I
havn't the energy to sort out what you are talking about.
Best to agree to differ, eh? ;o)
---
Reg.
W5DXP
> Dear Cec, your nitpicking is inexhaustible and at this time in the morning I
> havn't the energy to sort out what you are talking about.
>
> Best to agree to differ, eh? ;o)
I don't see how you can possibly assert that the fields are the same for
a standing-wave antenna and a traveling-wave antenna because they obviously
are not the same. A magnetic pickup at a current null on a standing-wave
antenna will read a low maximum value. A magnetic pickup at the same point
on a traveling-wave antenna will read a high maximum value.
Tdonaly
>
>I don't see how you can possibly assert that the fields are the same for
>a standing-wave antenna and a traveling-wave antenna because they obviously
>are not the same. A magnetic pickup at a current null on a standing-wave
>antenna will read a low maximum value. A magnetic pickup at the same point
>on a traveling-wave antenna will read a high maximum value.
>--
>73, Cecil http://www.qsl.net/w5dxp
>
What point are you talking about, Cecil? If I pick an arbitrary point on a
half-wave
dipole, how do I locate the same point on a traveling wave antenna?
73,
Tom Donaly, KA6RUH
W5DXP
> What point are you talking about, Cecil? If I pick an arbitrary point on a
> half-wave dipole, how do I locate the same point on a traveling wave antenna?
It's not me who is saying every 1/2WL is the same for both antennas
so I don't know how to locate the same point.
--
73, Cecil http://www.qsl.net/w5dxp
-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
tad danley
>
> An excellent and easy to follow description of the method of moments can
> be found in Kraus' _Antennas_, Second Ed. I assume it's in the third
> edition also, but it's not in the first. The NEC-2 manual recommends
> R.F. Harrington, _Field Computation by Moment Methods_ (McMillan, 1968)
> but I haven't seen this book.
I'm looking for a text to help me increase my understanding of antennas
beyond what is contained in the ARRL Antenna Handbook. It looks like
"Antennas" by Kraus is it. Can anyone recommend any others?
Thanks and 73,
--
* Do NOT use Reply *
Reply only through ARRL forwarding service to K3TD
Tad, K3TD
tad danley
>
> I've tried to point out on this thread that although the feedpoint
> impedance is an impedance with the units of ohms, and the impedance of a
> plane wave in free space also has the units of ohms, they're not the
> same thing.
This may not be a good analogy, but Specific Impulse of rocket motors
helps me to remember that the 'units' of something have to be considered
in the context of what is being measured. Specific impulse is a measure
of the performance of a rocket motor. It measures the thrust obtained
from a single kilogram of propellant burned in one second. The 'units'
of Specific Impulse are seconds, but we're not measuring 'time'.
W5DXP
> I'm looking for a text to help me increase my understanding of antennas
> beyond what is contained in the ARRL Antenna Handbook. It looks like
> "Antennas" by Kraus is it. Can anyone recommend any others?
_Antenna_Engineering_Handbook_, edited by Jasik, contributions by many.
_Antenna_Theory_Analysis_and_Design_, by Balanis
Richard Clark
wrote:
>Roy Lewallen wrote:
>
>>
>> I've tried to point out on this thread that although the feedpoint
>> impedance is an impedance with the units of ohms, and the impedance of a
>> plane wave in free space also has the units of ohms, they're not the
>> same thing.
>
>
>This may not be a good analogy, but Specific Impulse of rocket motors
>helps me to remember that the 'units' of something have to be considered
>in the context of what is being measured. Specific impulse is a measure
>of the performance of a rocket motor. It measures the thrust obtained
>from a single kilogram of propellant burned in one second. The 'units'
>of Specific Impulse are seconds, but we're not measuring 'time'.
>
>73,
Hi Tad,
Your point is well taken. ALL physical phenomenon can be expressed
through a chain of conversions in the MKS system of units. When
someone tells you that their terminology is inconsistent between
disciplines (as such offered in this and other threads); it must then
be amenable to reduction to MKS terms or one of the two conflicting
expressions is invalid.
That is to say to the specific matter about the usage of "ohms:"
Here, the unit of ohm must be reduced to Meters, Kilograms, and
Seconds for both usages (electrical and radiative). At that point,
both will have a common basis for comparison and if in fact their
reduced terms are identical, then their common usage is also
identical.
One simple example is with the measurement of body weight on the
bathroom scale (a torsion or compression device) as opposed to the
weight measured on the doctor's scale (a beam balance). Let's say
before you go to the doctor's, you weigh yourself in around 165
pounds. When you arrive at the doctor's, his scale says you weigh
around 75 kilograms.
Let's remove this same scenario to the moon (you live in one of those
futuristic 1990's colonies forecast by the space race back in the
60's). Before you went to the doctor's you weighed in around 33
pounds. When you arrive at the doctor's, his scale says you weigh
around 75 kilograms.
Here we find the expression "pounds" suffers what appears to be the
same plight of "ohms" in that the determination of a value is
inconsistent. You may also note constants of proportionality on earth
and the moon. These constants when expressed as a ratio also describe
the significant differences between the earth and the moon.
The problem is that the term "weight" has a hidden association to the
constant of Gravity. The expression Gram is one of Mass, not weight.
The expression pound is not an expression of Mass unless you expand it
to include the term for the particular constant of Gravity. Mass is
constant in the Newtonian Universe, and weight is not.
If you were to have reduced the pounds to the MKS system both times,
you would have found it consistent both times (here on earth, and on
the moon).
If you reduce the "ohms" to the MKS system both times.... Well I will
leave that for further deliberation as some are sure to be surprised.
:-)
Dr. Slick
>
> I've tried to point out on this thread that although the feedpoint
> impedance is an impedance with the units of ohms, and the impedance of a
> plane wave in free space also has the units of ohms, they're not the
> same thing. Feedpoint impedance is the ratio of a current to a voltage.
> Wave impedance, or the intrinsic impedance of a medium, is the ratio of
> an E field to an H field -- it's also the square root of the ratio of
> the medium's permeability to its permittivity. An antenna converts
> currents and voltages to E and H fields, it doesn't just transform one
> impedance to another. Hence my insistence on calling an antenna a
> transducer rather than a transformer.
>
I've agreed with you on the semantics of antennas as transducers,
but two transducers DO make a transformer.
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.
Slick
William E. Sabin
> Roy Lewallen wrote:
>
>>
>> An excellent and easy to follow description of the method of moments
>> can be found in Kraus' _Antennas_, Second Ed. I assume it's in the
>> third edition also, but it's not in the first. The NEC-2 manual
>> recommends R.F. Harrington, _Field Computation by Moment Methods_
>> (McMillan, 1968) but I haven't seen this book.
>
>
>
> I'm looking for a text to help me increase my understanding of antennas
> beyond what is contained in the ARRL Antenna Handbook. It looks like
> "Antennas" by Kraus is it. Can anyone recommend any others?
>
> Thanks and 73,
>
Kraus is not only an antenna expert, he is a
world-class authority on the entire field of
Electromagnetics, based on Maxwell's equations.
His mathematics is elegant.
Bill W0IYH
J. McLaughlin
Father, and knew where I wanted to go to grad school. He is also a very
fine person.
Buy and read his books.
73 Mac N8TT
--
J. Mc Laughlin - Michigan USA
Home: J...@Power-Net.Net
"William E. Sabin" <sabinw@mwci-news> wrote in message
news:vjvp8ro...@corp.supernews.com...
<snip>
Richard Clark
wrote:
>If you reduce the "ohms" to the MKS system both times.... Well I will
>leave that for further deliberation as some are sure to be surprised.
>:-)
Hi All,
from:
http://physics.nist.gov/cuu/Units/units.html
newton N m · kg · s^-2
volt V m^2 · kg · s^-3 · A^-1
farad F m^-2 · kg^-1 · s^4 · A^2
henry H m^2 · kg · s^-2 · A^-2
ohm omega m^2 · kg · s^-3 · A^-2
permittivity e0 farad per meter F/m
permeability µ0 henry per meter H/m
for e0 (electric constant) from:
http://physics.nist.gov/cgi-bin/cuu/Value?ep0|search_for=universal_in!
8.854187817 · 10^-12 · F · m^-1
for µ0 (magnetic constant) from:
http://physics.nist.gov/cgi-bin/cuu/Value?mu0|search_for=universal_in!
12.566370614 · 10^-7 · N · A^-2
Zc = (µ0 / e0)^0.5
solving:
Zc = 376.730 · ((N · A^-2) / (F · m^-1))^.5
expressed as products:
Zc = 376.730 · (N · A^-2 · F^-1 · m)^.5
substituting terms N and F:
Zc
=
376.730
·
((m · kg · s^-2 · A^-2)
·
(m^2 · kg · s^-4 · A^-2 * m))^.5
combining terms:
Zc = 376.730 · (m^4 · kg^2 · s^-6 · A^-4)^.5
root of terms:
Zc = 376.730 · m^2 · kg · s^-3 · A^-2
Which, if no one has noticed, is the same units for measuring the Ohm.
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. ;-)
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.
If someone has another method of deriving the Z of free space, they
may well rest assured that it will eventually collapse to the same
expression cited above. Offer your example for a similar tour through
the fascinating world of Physics.
You can still call it a transducer though - or a thigamajig.
Roy Lewallen
> . . .
> 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.
Sorry, that's a demonstrably absurd assertion. Tam pointed a good
example out a couple of days ago. Here's another -- The product of
resistance and capacitance is time. Do we "transform" time by running a
current through an RC circuit? There are countless other examples of
different quantities having the same units -- it's not hard to find them.
> If someone has another method of deriving the Z of free space, they
> may well rest assured that it will eventually collapse to the same
> expression cited above. Offer your example for a similar tour through
> the fascinating world of Physics.
That was lots of fun, but what does it have to do with the subject under
discussion.
> You can still call it a transducer though - or a thigamajig.
Gee, thanks!
Roy Lewallen, W7EL
Richard Clark
wrote:
>> . . .
>> 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.
>
>Sorry, that's a demonstrably absurd assertion.
Hi Roy,
>>although the feedpoint impedance is an impedance with the units of ohms,
>>and the impedance of a plane wave in free space also has the units of ohms,
>>they're not the same thing.
So, how do the "ohms" of free space differ from the "ohms" of a
quarter wave monopole's Radiation Resistance?
The demonstrables you offer do not enlighten us in what physical
constants these unique terms of your usage diverge from those in the
MKS system. As I pointed out in my posting, whatever derivation for
the characteristic Z of free space is, it must ultimately devolve to
the identical expression for the common Ohm.
I would offer by way of caution that the expression
Zc = (µ0 / e0)^0.5
where µ0 is expressed in Henrys per meter and
where e0 is expressed in Farads per meter;
that that, too, arrives at the same Ohms employed by carbon resistors
and Radiation Resistance.
This is much like trying to compare miles per gallon and kilometers
per liter. When push comes to shove, the reduction to MKS will reveal
that the same container of gas will get you down the same stretch of
asphalt the same distance no matter what mix of terms you substitute
for liquid volume and length. This goes to include offbeat
descriptive terms like miles per liter; kilometers per gallon; or
furlongs per hogshead for that identical container of gas. Solutions
of proportionality are not unique physical constants.
W5DXP
> You can still call it a transducer though - or a thigamajig.
Such a sophisticated concept deserves better. I suggest
"Triactuatedmulticomplicator" or TAMC for short.
Roy Lewallen
> . . .
> quarter wave monopole's Radiation Resistance?
As I and several others have recently said several times, the "ohms" of
free space represent a ratio of E field to H field of a plane wave or,
more fundamentally, the square root of the ratio of the medium's
permeability to its permittivity. The "ohms" of a quarter wave
monopole's radiation resistance represent the ratio of a voltage to a
current. An E field is not a voltage, and an H field is not a current.
They are different forms of energy, related by Maxwell's equations.
Roy Lewallen, W7EL
Richard Clark
wrote:
>Richard Clark wrote:
Hi Roy,
Physical constants do not betray physical measurements. The "ohms"
are identical in practice and theory. They are demonstrable in the
Standards laboratory. NIST describes all this at the links offered
and they do not equivocate nor banter terms casually. For any
Professional Engineer, they carry the force of law as the only
authoritative source for definition. I don't expect that standard to
be observed in this forum, as this is just hobby stuff and lord knows
the number of dictionaries that are already hauled off the shelf to
engage debate.
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.
Richard Clark
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. :-)
Dr. Slick
>
> 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
Dr. Slick
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 thing as a wave traveling through free space.
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.
Slick
Dr. Slick
>
> 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
k...@sympatico.ca
>
> 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
>
> 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
Richard Clark
>
>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.
Dr. 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.
>
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 here:
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
Richard Harrison
"Triactuatedmulticomplicator" or TAMC for short."
I have a suggested update:
"Triactuatedmultiuncomplicator", or TAMU for short.
Best regards, Richard Harrison, KB5WZI
k...@sympatico.ca
>
> k...@sympatico.ca wrote in message news:<3F41F6D7...@sympatico.ca>...
> > "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 here:
>
> 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
W5DXP
> Cecil, W5DXP wrote:
> "Triactuatedmulticomplicator" or TAMC for short."
>
> I have a suggested update:
> "Triactuatedmultiuncomplicator", or TAMU for short.
Richard, when I was there in the 50's, it was TAMC.
My '59 graduation ring says "A&M College of Texas".
Trivia note: At that time, Texas University was a
branch of the Texas A&M system. Gig 'Um!
Richard Harrison
"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.
Richard Harrison
"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."
Fields and the volts and amps they can produce are locked together. An
example is the 1-meter wire best placed to be swept by a radio wave of
1-millivolt per meter will have 1-millivolt induced in the wire.
The field is not identical to a voltage because it requires no conductor
to exist, but it has the same value and follows many of the same rules.
Roy Lewallen
(since 1981), I'm aware that I'm subject to state laws governing the
code of conduct of Professional Engineers, and all other applicable
state laws. I didn't realize that I had legal obligations to NIST, or
that any other federal agency has requirements for P.E.s of all states.
Would you please provide some reference where I can further research
this obligation and the rules it has imposed that I'm legally required
to comply with?
Roy Lewallen, W7EL, P.E.
Richard Clark wrote:
> . . .
> 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.
> . . .
Jim Kelley
> 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).
Torque is the same as work. Work is energy conversion.
> 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?
Yes. Sorta like knowing when it's legal to carry the one. ;-)
> What rules have you used to conclude that reducing V/m/A/m to V/A
> is appropriate?
Meters cancel meters.
I recently saw somewhere on the web where a given light intensity was
converted to the equivalent free space voltage and current. I think it
was a response to Dr. Slicks inquiry over on sci.physics.electromag.
....yes, here it is.
http://www.flashrock.com/upload/photong/photong.html
73 de ac6xg
Richard Clark
wrote:
>I find this most interesting. As a P.E. licensed by the state of Oregon
>(since 1981), I'm aware that I'm subject to state laws governing the
>code of conduct of Professional Engineers, and all other applicable
>state laws. I didn't realize that I had legal obligations to NIST, or
>that any other federal agency has requirements for P.E.s of all states.
>Would you please provide some reference where I can further research
>this obligation and the rules it has imposed that I'm legally required
>to comply with?
>
>Roy Lewallen, W7EL, P.E.
>
Hi Roy,
I am wholly unaware of the full scope of your business and contracts
and I have no interest, nor do I think you would volunteer that
information. I cannot recall a single instance of your relating any
experience of yours that revolved around the matters I have discussed,
nor any matters that were professional beyond your product. I cannot
imagine that your product enters into matters of traceability or
authority when I have seen your disclosures that explicitly remove
yourself from liability:
>Legal Disclaimer
>
>The licensee ("Licensee" or "User") acknowledges that the reliability
>of any and all results produced by this software are not precise and
>are subject to significant levels of variability.
...
>LICENSOR HEREBY DISCLAIMS ANY AND ALL WARRANTIES OF
>MERCHANTABILITY OR FITNESS FOR A PARTICULAR PURPOSE.
Roy Lewallen
mean to say that there are no such requirements? Or that you don't know
what they are?
Roy Lewallen, W7EL
Richard Clark
>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
Hi Keith,
Lets just observe a simple, real situation that any Ham may be faced
with during a power black-out, or during Field Day. Take for instance
a generator. It can give you 1KW of power. You need a gas powered
engine to turn the generator. How much horsepower do you need?
The common exchange is 746W per HP for 100% efficient transformation.
Thus you need at least 1.34 HP to obtain that kilowatt. What is a
horsepower (certainly one of the most ancient of units) compared to
these Watts (a relatively modern unit by comparison)? Is there a
direct correlation between the power of a horse, and the power of a
generator? Yes.
First, a word about multiplication by identities. An identity may
also be known in this forum as a conversion factor. One such simple
example is time conversion from seconds to minutes and back through:
(1 · minute) = (60 · second)
the identity is a simple division by one side or the other to leave 1.
A division by minute is a possibility for one identity:
1 = (60 · second) / (1 · minute)
equally valid would be to divide both original sides by (60 · second):
(1 · minute) / (60 · second) = 1
you can confirm there is no hanky-panky by observing the common
expectation that both sides of the equation describe the same thing,
thus the identity of (1) over (1) equals 1 --- both times. In other
words, the identity describes the same thing by different terms, and
those terms are combined to offer a value of 1 (dimensionless).
The process of employing multiplication by 1 (performed below) through
the use of identities with the time example described above (meaning
you have converted to a form of x = 1 or 1 = x) allows for us to
combine and clear terms in shifting from one basis of measurement to
another.
To return to our query about the generator and the engine,
1 Horsepower is 33,000 ft-lb/minute. In the old days, a horse had to
pull against a known load for a know period of time over a known
distance to arrive at this common reference. The popular definition
will allow you to see these units already in place:
33,000 · foot · pound / minute
We begin our trip towards the S of MKS through Units conversions, by
casting out minutes with the time identity multiplying this value:
33,000 · (foot · pound / minute) · (1 · minute) / (60 · s)
Clearing those terms leaves us with:
33,000 · foot · pound / (60 · s)
or
550 · foot · pound / s
when the minute terms are canceled and the equation has been corrected
to using seconds. [I hope many recognize this alternative conversion
factor. It proves that nothing is lost through these conversions.]
Next we move toward the K of MKS by casting out pounds:
550 · (foot · pound / s) · (1 · kg / 2.205 · pound)
This would be tempting to perform, but it would be absolutely wrong!
As far as the expression of power in the original statement goes, the
identity of pounds and kilograms is incorrect. This is because
kilograms express mass and pounds express weight, which is the product
of mass times the acceleration due to gravity. The pounds do cancel
in the equation above, but the statement is incomplete and should be:
550 · (foot · pound / s)
· (1 · kg / 2.205 · pound)
· (9.807 · m / s²)
Combining and casting out terms leaves us with:
2446 · foot · m · kg / s³
Finally, to complete the progress towards MKS, we move toward the M of
MKS by casting out foot using the length identity:
2446 · foot · m · kg / s³ · (0.3048 · m) / (1 · foot)
Combining and clearing terms leaves us with:
745.5 · m² · kg / s³
THIS is the NIST definition for power, but as such it may be
unfamiliar to many (certainly given the angst and denial that attends
this discussion). For the comfort of many, we draw in another
identity that comes closer to expectations.
That is the identity of Power (also in MKS terms) that reveals itself
as joules per second, or newton-meters per second:
(1 · Watt) = (1 · kg · m / s²) · (m) / (s)
or
(1 · Watt) = (1 · kg · m² / s³)
whose identity becomes
(1 · Watt) / (1 · kg · m² / s³) = 1
We apply this to the power equation above:
745.5 · (m² · kg / s³) · (Watt) / (kg · m² / s³)
which (guess what?) reduces to:
745.5 Watts
QED
Rounding introduced 0.5 Watt error (the values provided by NIST to
their complete precision would eliminate that). It also confirms what
we already knew, but few could prove with a linear exercise like this.
That's not uncommon however, because few deal with the Physics of the
terms they are familiar with, this is the provence of the Metrologist
and research scientists, not amateurs.
It is enough to say Watts and Horse Power exhibit a constant of
proportionality, but it is wholly wrong to say that electrical Watts
are somehow different from an animal's work expended over time.
It is equally in error to maintain that the resistance or Z of free
space is somehow remote and different from the resistance of a carbon
composition resistor or Radiation Resistance. ALL terms employed in
the expression of permittivity and permeability conform to these same
linear operations that prove they are congruent.
k...@sympatico.ca
You have picked an example that simply has different representations
for power. I do not believe there has been any dispute about whether
conversions between different units of power are valid; they are.
The general question is: if two things can be simplified to the same
set of units are they the same thing.
At least two counter examples have been offerred to demonstrate that
just because two things have the same units, they are not the same.
Torque is not work; though they both have N-m as their units.
Modulus of elasticity is not stress; though they are both expressed
as Pascals (after simplification).
This seems sufficient to prove that two things with the same units
are not necessarily the same.
It leaves open the question as to how does one know whether two
things with the same units are the same (or not); a much more
challenging problem, I suspect.
...Keith
Dr. Slick
> >
> > 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 ...
>
But it's still dimensionless, a "pure number" as they call it.
> So dimensionless quantities are not all the same, even though they are
> all dimensionless.
>
In context, i would agree, but they are still just pure numbers.
> >
> > 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.
>
ok, length of angle divided by length of radius, that's right.
> After multiplying Torque by Radians, you have computed the length
> along the arc through which the force has acted -> energy, of course.
>
>
> Actually, you've done 2*pi*radius*force work. Moving one circumference
> times the force.
>
Actually, thats 2*pi*radius*force*moment arm. Right.
> > > ...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.
>
Your point it well taken, but cancellation of units has always
worked for me, and everyone else i went to college and high school
with.
And I admit that a wave traveling in a transmission line is
different from an EM wave traveling through space. All i'm saying is
that the E field is defined by a voltage potential field, and the H
field by amperes, so to say that the E field has nothing to do with
voltage potential is a wrong statement in my opinion. And it's still
ohms for the impedance.
> 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
It's a difficult question, and i'm glad we are discussing it. All
i'm saying is, the units have to be the same, where ever you use them.
Slick
Richard Clark
>Good day Richard,
>
>You have picked an example that simply has different representations
>for power. I do not believe there has been any dispute about whether
>conversions between different units of power are valid; they are.
>
>The general question is: if two things can be simplified to the same
>set of units are they the same thing.
>At least two counter examples have been offerred to demonstrate that
>just because two things have the same units, they are not the same.
>
>Torque is not work; though they both have N-m as their units.
If you take a solid axle, fix it at one end and twist at the other,
Torque is the plastic deformation in the form of that twist being
distributed along the length of the axle as shearing stress. That
twist allows for some rotation at the end where the rotational force
is applied and that is obviously work. I know, I've calibrated 100's
of Torque wrenches (mostly micrometer click wrenches) from 15 pound-in
to a 600 pound-ft and broke a bench doing it.
>Modulus of elasticity is not stress; though they are both expressed
>as Pascals (after simplification).
This has a close association with your observation above, so I will
continue with the same model. But first, the definitions that you
seem to accept, but tied into this discussion. From "University
Physics," Sears and Zemansky, containing a chapter called "Elasticity"
whose second section is titled "Stress" (the first section is titled
"Introduction").
"Stress is a force per unit area."
"Strain. ...refers to the relative change in dimensions ... subjected
to stress." As this is distance over distance, strain has no
dimension (the units cancel as has been pointed out by others).
"Elastic modulus. The ratio of a stress to the corresponding strain
is called an elastic modulus. ... Since a strain is a pure number,
the units of Young's modulus are the same as those of stress, namely,
force per unit area. Tabulated values are usually in lb/in² or
dynes/cm²."
Returning to that same axle. We score a line along its length from
free end to fixed end with a scribe that travels a path parallel to
the axis. We apply some force, hold it, and scribe a second line. We
go to the middle of its length and scribe two lines around the
circumference of the axle (a short distance apart). These last two
lines describe opposite shears due to torsion. The stress varies as a
function of depth into the axle (greater at the periphery, less in the
interior). We then examine the enclosed area which describes a twist
per length (area for the applied force - stress). The axial lines are
parallel to the compression and the circumferential lines are parallel
to the tension. If this axle were made of wood, it would fail under
compression when its elasticity was pushed beyond its limit. In
comparison, it would also exhibit a larger rotational displacement
with the same force applied to an iron axle.
The last observation is simply reduced, or normalized as described
above in the definition of modulus and what you obtain for the two
materials is either a constant force with different rotational
displacements (and different scribed areas); or the same rotational
displacement (constant scribed areas) with different applied forces.
You still have torsion, you still have stress and strain, and you
still have rotational displacement - the only difference is in the
material's characteristic which is described by the modulus.
>
>This seems sufficient to prove that two things with the same units
>are not necessarily the same.
It proves you have two different materials which is the point of a
modulus in any discipline.
>
>It leaves open the question as to how does one know whether two
>things with the same units are the same (or not); a much more
>challenging problem, I suspect.
>
>...Keith
Hi Keith,
The only question is what is different, the why follows from fairly
obvious implications of being material based. Your problem is in the
definition of the application of the terms, not their expressions.
You might want to consult a slim volume called
"Elements of Strength of Materials," Timoshenko.
You will note that this bears no relation to ohms being different,
because as you observed with the horsepower example, it is simply
flipping through translations until you hit the units you want.
k...@sympatico.ca
> > times the force.
> >
>
> Actually, thats 2*pi*radius*force*moment arm. Right.
In my example, I intended the 'radius' to be the radius at which the
force was applied so the 'moment arm' was already accounted for. When
the radius is the radius at which the force is applied, 2*pi*radius
is the distance through which the force has acted after one revolution
so the expression is the same as the common force*distance used for
linear work. More generally, it does not matter what the shape of the
path is; the work is always the force times the distance along the path.
...Keith
Dr. Slick
So then that would be Work = 2*pi*(radius**2)*force.
Right. But it's a bit strange with rotational work, because if
you do one rotation, you are essentially back where you started. I
believe this has something to do with the fact that the rotational
displacement is dimensionless, and linear displacement is in units of
length.
Slick
Dr. Slick
>
> >
> >It leaves open the question as to how does one know whether two
> >things with the same units are the same (or not); a much more
> >challenging problem, I suspect.
> >
> >...Keith
>
>
> because as you observed with the horsepower example, it is simply
> flipping through translations until you hit the units you want.
>
> 73's
> Richard Clark, KB7QHC
I don't think anyone here is arguing that a wave traveling
through a transmission line is the same as an EM wave traveling
through free-space.
But as Richard has shown, the units are always the same, as they
should be. Just like a meter is still a meter, whether it is in
torque or work.
But it tells you something about what you are measuring, and the
clue is that the E field is defined by the voltage potential field,
and the H field by amps (turns).
And if the permittivity (impedance) of the material surrounding
an antenna will affect it's input impedance, i think it's something to
consider.
Slick
k...@sympatico.ca
>
> So then that would be Work = 2*pi*(radius**2)*force.
>
> Right. But it's a bit strange with rotational work, because if
> you do one rotation, you are essentially back where you started. I
> believe this has something to do with the fact that the rotational
> displacement is dimensionless, and linear displacement is in units of
> length.
If it takes one newton to make a block move and I push it north 1
metre, then east 1 metre, then south 1 metre, then west 1 metre,
I am back where I started but I did 4 joules worth of work (but
probably not useful work).
If I push the same block in a circle of radius 1 metre then I do
2pi joules of work: the circumference (distance travelled) times
the force.
So the expression at the top should not have a square in it.
...Keith
Dr. Slick
> "Dr. Slick" wrote:
> >
> > So then that would be Work = 2*pi*(radius**2)*force.
> >
> > Right. But it's a bit strange with rotational work, because if
> > you do one rotation, you are essentially back where you started. I
> > believe this has something to do with the fact that the rotational
> > displacement is dimensionless, and linear displacement is in units of
> > length.
>
> If it takes one newton to make a block move and I push it north 1
> metre, then east 1 metre, then south 1 metre, then west 1 metre,
> I am back where I started but I did 4 joules worth of work (but
> probably not useful work).
>
Ok, this is what i meant, "usefull work".
> If I push the same block in a circle of radius 1 metre then I do
> 2pi joules of work: the circumference (distance travelled) times
> the force.
>
> So the expression at the top should not have a square in it.
>
> ...Keith
Nope. It needs another radius term...one for the circumference,
and one for the moment arm.
Slick
k...@sympatico.ca
Energy = Force x Distance
When I push the block in a square pattern, the distance is the sum of
the sides of the square.
When I push the block in a circle, the distance is the circumference,
which is 2 x pi x Radius.
If, after reviewing the above, you still think the radius needs
squaring, please explain why the equation for energy when pushing
in a circle is different than that for pushing in a square.
As an exercise, consider pushing in a triangle pattern, a square
pattern, pentagon, hexagon, heptagon, octagon, ....
Each pattern above is getting closer and closer to being a circle;
at which pattern does the equation for Energy change from being
the distance moved to being twice the enclosed area?
...Keith
k...@sympatico.ca
>
> Please think carefully about your reply.
>
> Energy = Force x Distance
>
> When I push the block in a square pattern, the distance is the sum of
> the sides of the square.
>
> When I push the block in a circle, the distance is the circumference,
> which is 2 x pi x Radius.
>
> If, after reviewing the above, you still think the radius needs
> squaring, please explain why the equation for energy when pushing
> in a circle is different than that for pushing in a square.
>
> As an exercise, consider pushing in a triangle pattern, a square
> pattern, pentagon, hexagon, heptagon, octagon, ....
>
> Each pattern above is getting closer and closer to being a circle;
> at which pattern does the equation for Energy change from being
> the distance moved to being twice the enclosed area?
And I forgot to say...
In a rotary system, it is often deemed convenient to think of Distance
in terms of rotations or angular displacement.
Torque is just a way of thinking about Force so that you can use
angular displacement instead of distance.
...Keith
Richard Harrison
"And if the permittivity (impedance) of the material surrounding an
consider."
The permittivity surrounding our antennas rarely changes and is the same
for nearly all antennas.
My dictionary says of permittivity: "See Dielectric Constant".
Velocity can be affected by dielectric constant as is seen in
solid-dielectric coax. Fortunately, the dielectric constant of the
environment our antennas operate in is nearly constant.
Were matching antennas to 377 ohms significant, it would manifest itself
in the century of experience of using many antennas of many differing
types.
Dr. Slick
> Please think carefully about your reply.
>
I always do, unless i hafta take a dump! hehe..
> Energy = Force x Distance
>
> When I push the block in a square pattern, the distance is the sum of
> the sides of the square.
>
> When I push the block in a circle, the distance is the circumference,
> which is 2 x pi x Radius.
>
> If, after reviewing the above, you still think the radius needs
> squaring, please explain why the equation for energy when pushing
> in a circle is different than that for pushing in a square.
>
> As an exercise, consider pushing in a triangle pattern, a square
> pattern, pentagon, hexagon, heptagon, octagon, ....
>
> Each pattern above is getting closer and closer to being a circle;
> at which pattern does the equation for Energy change from being
> the distance moved to being twice the enclosed area?
>
> ...Keith
>
It looks like you are correct. Sorry for the mistake, it's been
too
long since i've done a torque problem! I'm more of a EE!
Look at this page:
http://www.sinclair.net/~ddavis/170_ps10.html
If you agree that torque is in units of N*m, then according to this
page, you have to multiply this by the angular displacement, which
will be 2*pi for a full revolution. I was incorrectly trying to
multiple by the actual circumference traveled, which is incorrect
because we want angular displacement instead.
To me, this kinda shows how torque is closer to work than just
force, because with torque, you just need to and the angular
displacement (dimensionless) to get the work done.
Thanks for the review, Keith.
Slick
Dr. Slick
I agree with you that almost all antennas are surrounded by air or
something very close to free-space (outer-space).
Many have agreed with me that a water-proof antenna emmersed in
water will change in it's input impedance, because of a different
dielectric constant.
I'm not sure how many antennas are designed for underwater use,
but it doesn't matter for the sake of arguing that the antenna
"matches" to the impedance of the material that surrounds it.
Slick
Richard Clark
>
>As an exercise, consider pushing in a triangle pattern, a square
>pattern, pentagon, hexagon, heptagon, octagon, ....
>
>Each pattern above is getting closer and closer to being a circle;
>at which pattern does the equation for Energy change from being
>the distance moved to being twice the enclosed area?
>
>...Keith
>
Hi All,
Just what purpose do the two of you think you are achieving with
Torque and boxing the compass?
Tdonaly
>Hi All,
>
>Just what purpose do the two of you think you are achieving with
>Torque and boxing the compass?
>
>73's
>Richard Clark, KB7QHC
Circle squarers are ten times worse than flat-earthers,
turtles or not.
73,
Tom Donaly, KA6RUH
Richard Clark
wrote:
>I find this most interesting. As a P.E. licensed by the state of Oregon
>(since 1981), I'm aware that I'm subject to state laws governing the
>code of conduct of Professional Engineers, and all other applicable
>state laws. I didn't realize that I had legal obligations to NIST, or
>that any other federal agency has requirements for P.E.s of all states.
>Would you please provide some reference where I can further research
>this obligation and the rules it has imposed that I'm legally required
>to comply with?
>
>Roy Lewallen, W7EL, P.E.
Hi Roy,
"RCW 19.94.150
Standards recognized.
The system of weights and measures in customary use in the United
States and the metric system of weights and measures are jointly
recognized, and either one or both of these systems shall be used
for all commercial purposes in this state. The definitions of
basic units of weight and measure and weights and measures
equivalents, as published by the national institute of standards
and technology or any successor organization, are recognized and
shall govern weighing or measuring instruments or devices used in
commercial activities and other transactions involving weights and
measures within this state."
This is from the state of Washington, I will leave it to you to
research your own particular point of liability in Oregon.
I would add what the IEEE offers into the matter of observing
standards in the development of software and confirming your
disclaimers with:
"The Legal Standard of Professionalism"
"One curious fact from the legal perspective decries a serious
lack: there is no such thing as software malpractice. Why?
A peek into the legal mind provides a disturbing explanation.
There is insufficient evidence to show that programmers
know how to learn from each other, much less from the rest of
the world."
I, for one, could envision you having interest in both, but as I
stated before, I could not see you bothered with the first - seeing
that you have not volunteered any additional details of your trade
aside from software, that stands to good reason.
Roy Lewallen
Wow, thanks for the heads-up. I'll be more careful to specify circuit
board trace line widths in furlongs, and volumes of radar detection
regions in bushels, those being duly recognized customary units of
measure here in Oregon. I'll no longer use lakj;ofs and mapeurqak!pys,
which I had previously been using.
> I would add what the IEEE offers into the matter of observing
> standards in the development of software and confirming your
> disclaimers with:
>
> "The Legal Standard of Professionalism"
>
> "One curious fact from the legal perspective decries a serious
> lack: there is no such thing as software malpractice. Why?
> A peek into the legal mind provides a disturbing explanation.
> There is insufficient evidence to show that programmers
> know how to learn from each other, much less from the rest of
> the world."
>
> I, for one, could envision you having interest in both, but as I
> stated before, I could not see you bothered with the first - seeing
> that you have not volunteered any additional details of your trade
> aside from software, that stands to good reason.
As I'm afraid so often happens with your postings, I haven't a clue what
you're trying to say. It sounds vaguely like a complaint, but I can't
for the life of me fathom what about, except that it seems to be some
sort of objection to the legal disclaimers which accompany my software.
Could you please try to rephrase it in a way that can be understood by
an engineer with a sadly deficient liberal arts education?
If you feel that the legal disclaimers which accompany my software are
unduly restrictive or otherwise too onerous for you, or you're not fully
satisfied with EZNEC in any way, all you need do is so state in
peasant-level plain language so I can understand it, and I'll promptly
refund the full purchase price. Just as it says clearly in the EZNEC
manual (Help/Contents/Introduction/Guarantee).
Roy Lewallen, W7EL
Richard Clark
wrote:
Uh-huh.
...
>Could you please try to rephrase it in a way that can be understood by
>an engineer with a sadly deficient liberal arts education?
Hi Roy,
Probably not.
Roy Lewallen
Surely, then, one of the more educated but earthy readers understood it
and can translate for me. Anyone? Here it is again in case it was missed
the first time:
----- Text to translate:
I would add what the IEEE offers into the matter of observing
standards in the development of software and confirming your
disclaimers with:
"The Legal Standard of Professionalism"
"One curious fact from the legal perspective decries a serious
lack: there is no such thing as software malpractice. Why?
A peek into the legal mind provides a disturbing explanation.
There is insufficient evidence to show that programmers
know how to learn from each other, much less from the rest of
the world."
I, for one, could envision you having interest in both, but as I
stated before, I could not see you bothered with the first - seeing
that you have not volunteered any additional details of your trade
aside from software, that stands to good reason.
------
What's the point? Can someone clue me in?
Roy Lewallen, W7EL
Gene Nygaard
wrote:
>On Sun, 17 Aug 2003 13:56:11 GMT, tad danley <nos...@atsbcglobal.net>
>wrote:
>
>>Roy Lewallen wrote:
>>
>>>
>>> I've tried to point out on this thread that although the feedpoint
>>> impedance is an impedance with the units of ohms, and the impedance of a
>>> plane wave in free space also has the units of ohms, they're not the
>>> same thing.
>>
>>
>>This may not be a good analogy, but Specific Impulse of rocket motors
>>helps me to remember that the 'units' of something have to be considered
>>in the context of what is being measured. Specific impulse is a measure
>>of the performance of a rocket motor. It measures the thrust obtained
>>from a single kilogram of propellant burned in one second. The 'units'
>>of Specific Impulse are seconds, but we're not measuring 'time'.
>>
>>73,
>
>Hi Tad,
>
>Your point is well taken. ALL physical phenomenon can be expressed
>through a chain of conversions in the MKS system of units. When
>someone tells you that their terminology is inconsistent between
>disciplines (as such offered in this and other threads); it must then
>be amenable to reduction to MKS terms or one of the two conflicting
>expressions is invalid.
>
>That is to say to the specific matter about the usage of "ohms:"
>Here, the unit of ohm must be reduced to Meters, Kilograms, and
>Seconds for both usages (electrical and radiative). At that point,
>both will have a common basis for comparison and if in fact their
>reduced terms are identical, then their common usage is also
>identical.
>
>One simple example is with the measurement of body weight on the
>bathroom scale (a torsion or compression device) as opposed to the
>weight measured on the doctor's scale (a beam balance). Let's say
>before you go to the doctor's, you weigh yourself in around 165
>pounds. When you arrive at the doctor's, his scale says you weigh
>around 75 kilograms.
>
>Let's remove this same scenario to the moon (you live in one of those
>futuristic 1990's colonies forecast by the space race back in the
>60's). Before you went to the doctor's you weighed in around 33
>pounds. When you arrive at the doctor's, his scale says you weigh
>around 75 kilograms.
>
>Here we find the expression "pounds" suffers what appears to be the
>same plight of "ohms" in that the determination of a value is
>inconsistent. You may also note constants of proportionality on earth
>and the moon. These constants when expressed as a ratio also describe
>the significant differences between the earth and the moon.
>
>The problem is that the term "weight" has a hidden association to the
>constant of Gravity. The expression Gram is one of Mass, not weight.
>The expression pound is not an expression of Mass unless you expand it
>to include the term for the particular constant of Gravity. Mass is
>constant in the Newtonian Universe, and weight is not.
>
>If you were to have reduced the pounds to the MKS system both times,
>you would have found it consistent both times (here on earth, and on
>the moon).
Apparently you are claiming that pounds are not units of mass.
Where did you learn that?
Being the skeptic that I am, how can I convince myself that that is
true? Is there some textbook, or something from some national
standards agency, that would help me verify this?
Gene Nygaard
Gene Nygaard
wrote:
>Roy Lewallen wrote:
>
>>
>> I've tried to point out on this thread that although the feedpoint
>> impedance is an impedance with the units of ohms, and the impedance of a
>> plane wave in free space also has the units of ohms, they're not the
>> same thing.
>
>
>This may not be a good analogy, but Specific Impulse of rocket motors
>helps me to remember that the 'units' of something have to be considered
>in the context of what is being measured. Specific impulse is a measure
>of the performance of a rocket motor. It measures the thrust obtained
>from a single kilogram of propellant burned in one second. The 'units'
>of Specific Impulse are seconds, but we're not measuring 'time'.
It is a bad analogy--for the simple fact that in SI, the proper units
of specific impulse are newton seconds per kilogram (N·s/kg), or the
equivalent meters per second (m/s).
Gene Nygaard
http://ourworld.compuserve.com/homepages/Gene_Nygaard/
Richard Clark
wrote:
>
>Apparently you are claiming that pounds are not units of mass.
>
>Where did you learn that?
>
>Being the skeptic that I am, how can I convince myself that that is
>true? Is there some textbook, or something from some national
>standards agency, that would help me verify this?
>
>Gene Nygaard
Hi Gene,
Exactly. Perhaps you should re-consider the simple illustration of
difference that I offered in the post you responded to.
Does the weight you measure on a bathroom scale change from the earth
to the moon because your mass changed too? Jenny Craig would have an
armada of shuttles warming up in Florida to a steady trade if that
were true.
However, you do ask for a reference and acknowledge the NIST as a
reputable source (many here ignore this commonplace):
http://physics.nist.gov/PhysRefData/contents-constants.html
The link:
http://physics.nist.gov/cuu/Units/units.html
is quite specific to the matter.
One of the supreme ironies comes in the form of the unstated
conditional. In your regard, it is pounds is intimately tied to the
gravitational constant (mass and G). In other regards SWR is
intimately tied to the source Z (always equal to the transmission line
characteristic Z, unless stated otherwise).
Dave Shrader
I better stay here. But my doctor wants me to loose 40#. Something's
wrong. I need a Twinkie!!
DD
Richard Clark
<david....@comcast.net> wrote:
>I weigh approximately 40# on the moon!!! Too skinny for my height 5"9".
>
>I better stay here. But my doctor wants me to loose 40#. Something's
>wrong. I need a Twinkie!!
>
>DD
Hi OM,
If you huff down a package of Ex-Lax you would take care of the
doctor's advice with a lot of "loose" weight. (Language is fun ;-)
Gene Nygaard
wrote:
>On Wed, 24 Sep 2003 16:45:38 GMT, Gene Nygaard <gnyg...@nccray.com>
>wrote:
>
>>
>>Apparently you are claiming that pounds are not units of mass.
>>
>>Where did you learn that?
>>
>>Being the skeptic that I am, how can I convince myself that that is
>>true? Is there some textbook, or something from some national
>>standards agency, that would help me verify this?
>>
>>Gene Nygaard
>
>Hi Gene,
>
>Exactly. Perhaps you should re-consider the simple illustration of
>difference that I offered in the post you responded to.
>
>Does the weight you measure on a bathroom scale change from the earth
>to the moon because your mass changed too? Jenny Craig would have an
>armada of shuttles warming up in Florida to a steady trade if that
>were true.
So what happens when you get serious about your weight and go to the
doctors office or the gym and weigh yourself on one of those platform
type beam balances?
Would your pounds be different on the moon? By how much?
>However, you do ask for a reference and acknowledge the NIST as a
>reputable source (many here ignore this commonplace):
>http://physics.nist.gov/PhysRefData/contents-constants.html
There is absolutely nothing about pounds on this page. So don't be
bullshitting us.
>The link:
>
>is quite specific to the matter.
Not a link directly on the page above; maybe on one of the links
there.
There is absolutely nothing about pounds on this page either. You are
still bullshittiing.
>One of the supreme ironies comes in the form of the unstated
>conditional. In your regard, it is pounds is intimately tied to the
>gravitational constant (mass and G).
They are? I asked you for some citation proving that pounds are not
units of mass. You have not done so.
--
Gene Nygaard
http://ourworld.compuserve.com/homepages/Gene_Nygaard/
"It's not the things you don't know
what gets you into trouble.
"It's the things you do know
that just ain't so."
Will Rogers