Discussion:
Supporting theory that Antennas "Match" to 377 Ohms (Free space)
(too old to reply)
Dr. Slick
2003-08-08 11:21:48 UTC
Permalink
Hi,

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)
2003-08-08 14:09:01 UTC
Permalink
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.
--
73 es cul

wb3fup
a Salty Bear
Post by Dr. Slick
Hi,
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
Richard Clark
2003-08-08 18:53:31 UTC
Permalink
On Fri, 8 Aug 2003 10:09:01 -0400, "WB3FUP \(Mike Hall\)"
Post by WB3FUP (Mike Hall)
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
Reg Edwards
2003-08-08 20:17:17 UTC
Permalink
Tricky Dick Sez -
Post by Richard Clark
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.
Tricky,
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
Yuri Blanarovich
2003-08-08 14:46:35 UTC
Permalink
Post by Dr. Slick
Hi,
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
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
2003-08-08 15:00:38 UTC
Permalink
Dr. Slick wrote:
"---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
W5DXP
2003-08-08 16:10:52 UTC
Permalink
Post by Richard Harrison
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
Reg Edwards
2003-08-08 15:35:51 UTC
Permalink
An antenna's radiating efficiency has nothing whatever to do with the
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
2003-08-08 16:18:40 UTC
Permalink
Post by Reg Edwards
An antenna's radiating efficiency has nothing whatever to do with the
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.
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
Reg Edwards
2003-08-08 18:23:14 UTC
Permalink
Post by W5DXP
Post by Reg Edwards
Now I suppose somebody will drag in the irrelevant matter of SWR on the
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.
W5DXP
2003-08-08 19:08:51 UTC
Permalink
Post by Reg Edwards
Post by 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? :-)
Post by Reg Edwards
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
2003-08-08 19:58:09 UTC
Permalink
Post by W5DXP
I am much more interested in getting the
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!
Dave Platt
2003-08-08 21:01:16 UTC
Permalink
Post by Reg Edwards
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 <***@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!
Roy Lewallen
2003-08-09 19:36:40 UTC
Permalink
Only at some distance from the antenna. You can create local E/H ratios
of nearly any value (magnitude and phase).

Roy Lewallen, W7EL
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).
William E. Sabin
2003-08-09 20:32:01 UTC
Permalink
Post by Roy Lewallen
Only at some distance from the antenna. You can create local E/H ratios
of nearly any value (magnitude and phase).
Roy Lewallen, W7EL
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).
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
2003-08-09 20:57:17 UTC
Permalink
Post by Roy Lewallen
Only at some distance from the antenna. You can create local E/H
ratios of nearly any value (magnitude and phase).
Roy Lewallen, W7EL
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).
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
Richard Clark
2003-08-10 16:00:10 UTC
Permalink
On Sun, 10 Aug 2003 08:54:56 -0500, "William E. Sabin"
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. ;-)

73's
Richard Clark, KB7QHC
William E. Sabin
2003-08-10 19:28:11 UTC
Permalink
Post by Richard Clark
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
2003-08-10 20:28:47 UTC
Permalink
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.) 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
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
David Robbins
2003-08-10 21:42:37 UTC
Permalink
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.
Post by Roy Lewallen
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.) 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
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
Dr. Slick
2003-08-11 01:21:27 UTC
Permalink
Post by David Robbins
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
Roy Lewallen
2003-08-10 22:07:35 UTC
Permalink
No, it's really more a matter of how the antenna is oriented relative to
the flow of the Earth's Chi.

Roy Lewallen, W7EL
Post by Roy Lewallen
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
William E. Sabin
2003-08-15 14:40:12 UTC
Permalink
Post by Roy Lewallen
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
Post by Roy Lewallen
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
W5DXP
2003-08-15 15:05:36 UTC
Permalink
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
2003-08-15 17:21:29 UTC
Permalink
Post by 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?
===============================

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.
Reg Edwards
2003-08-16 00:11:49 UTC
Permalink
How can I lead myself astray by asking a question? :-)
------------------------------------------------------------------
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
2003-08-16 01:23:10 UTC
Permalink
Post by Reg Edwards
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



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Reg Edwards
2003-08-16 01:51:07 UTC
Permalink
Post by W5DXP
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.
============================

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
2003-08-16 02:16:23 UTC
Permalink
Post by Reg Edwards
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.
--
73, Cecil http://www.qsl.net/w5dxp



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Roy Lewallen
2003-08-15 18:06:29 UTC
Permalink
It's a simple matter to model a Beverage with EZNEC and observe the near
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
Post by 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?
--
73, Cecil, W5DXP
Richard Clark
2003-08-15 17:39:32 UTC
Permalink
On Fri, 15 Aug 2003 09:40:12 -0500, "William E. Sabin"
Post by William E. Sabin
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.

73's
Richard Clark, KB7QHC
William E. Sabin
2003-08-15 19:57:46 UTC
Permalink
Post by William E. Sabin
Post by Roy Lewallen
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.
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
2003-08-15 20:38:47 UTC
Permalink
On Fri, 15 Aug 2003 14:57:46 -0500, "William E. Sabin"
Post by William E. Sabin
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.

73's
Richard Clark, KB7QHC
Roy Lewallen
2003-08-15 21:13:06 UTC
Permalink
EZNEC doesn't do the transformation you describe.

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
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
tad danley
2003-08-17 13:42:10 UTC
Permalink
Post by Roy Lewallen
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
W5DXP
2003-08-17 14:28:07 UTC
Permalink
Post by tad danley
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
--
73, Cecil http://www.qsl.net/w5dxp



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William E. Sabin
2003-08-17 20:27:31 UTC
Permalink
Post by tad danley
Post by Roy Lewallen
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
2003-08-17 21:32:13 UTC
Permalink
...and his writing is lucid. I read his first edition, a gift from my
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: ***@Power-Net.Net

"William E. Sabin" <***@mwci-news> wrote in message news:***@corp.supernews.com...
<snip>
Post by William E. Sabin
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
Richard Clark
2003-08-17 17:14:21 UTC
Permalink
Post by Roy Lewallen
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.
:-)

73's
Richard Clark, KB7QHC
Richard Clark
2003-08-18 19:13:39 UTC
Permalink
. . .
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.

73's
Richard Clark, KB7QHC
Dr. Slick
2003-08-19 03:03:48 UTC
Permalink
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
2003-08-19 10:07:19 UTC
Permalink
Post by 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.
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
Richard Clark
2003-08-19 16:44:02 UTC
Permalink
Post by k***@sympatico.ca
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.
Post by k***@sympatico.ca
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)
Post by k***@sympatico.ca
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.
Post by k***@sympatico.ca
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
Post by k***@sympatico.ca
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.
Post by k***@sympatico.ca
What rules have you used to conclude that reducing V/m/A/m to V/A
is appropriate?
...Keith
Hi Keith,

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

From page 3 (yes, pretty up front):

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

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

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

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

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

73's
Richard Clark, KB7QHC
Roy Lewallen
2003-08-19 21:47:58 UTC
Permalink
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.
Post by Richard Clark
. . .
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.
. . .
Richard Clark
2003-08-19 23:58:22 UTC
Permalink
Post by Roy Lewallen
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
Post by Roy Lewallen
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.
...
Post by Roy Lewallen
LICENSOR HEREBY DISCLAIMS ANY AND ALL WARRANTIES OF
MERCHANTABILITY OR FITNESS FOR A PARTICULAR PURPOSE.
73's
Richard Clark, KB7QHC
Richard Clark
2003-08-22 02:35:48 UTC
Permalink
Post by Roy Lewallen
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.

73's
Richard Clark, KB7QHC
Roy Lewallen
2003-08-22 03:03:41 UTC
Permalink
Post by Richard Clark
Post by Roy Lewallen
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.
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.
Post by Richard Clark
I would add what the IEEE offers into the matter of observing
standards in the development of software and confirming your
"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
2003-08-22 03:25:17 UTC
Permalink
Post by Roy Lewallen
Post by Richard Clark
Post by Roy Lewallen
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.
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.
Uh-huh.

...
Post by Roy Lewallen
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.

73's
Richard Clark, KB7QHC
Roy Lewallen
2003-08-22 03:49:35 UTC
Permalink
Post by Richard Clark
...
Post by Roy Lewallen
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.
73's
Richard Clark, KB7QHC
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
Dr. Slick
2003-08-19 17:13:27 UTC
Permalink
Post by k***@sympatico.ca
Post by 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.
How do you know when the reduced units of one computation mean the
same thing as another?
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.
Post by k***@sympatico.ca
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.
Post by k***@sympatico.ca
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
k***@sympatico.ca
2003-08-19 18:35:55 UTC
Permalink
Post by Dr. Slick
Post by k***@sympatico.ca
Post by 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.
How do you know when the reduced units of one computation mean the
same thing as another?
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.
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.
Post by Dr. Slick
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.
Post by k***@sympatico.ca
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.
Post by Dr. Slick
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.
Post by Dr. Slick
But the crux is
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.
Post by k***@sympatico.ca
So sometimes it is appropriate to say the reduced results are the
same and some times it is not. Is there a way to know when it is
legal?
What rules have you used to conclude that reducing V/m/A/m to V/A
is appropriate?
...Keith
Basic algebra and cancellation of units. When have you found it
not to be appropriate?
It is not appropriate to consider Torque and Work to be the same, though
they have the same units.

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

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

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

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

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

...Keith
Richard Clark
2003-08-20 02:30:42 UTC
Permalink
Post by k***@sympatico.ca
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.

73's
Richard Clark, KB7QHC
k***@sympatico.ca
2003-08-20 03:34:07 UTC
Permalink
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.
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
Post by Richard Clark
Post by k***@sympatico.ca
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
(1 · minute) = (60 · second)
the identity is a simple division by one side or the other to leave 1.
1 = (60 · second) / (1 · minute)
(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
33,000 · foot · pound / minute
We begin our trip towards the S of MKS through Units conversions, by
33,000 · (foot · pound / minute) · (1 · minute) / (60 · s)
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.]
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
550 · (foot · pound / s)
· (1 · kg / 2.205 · pound)
· (9.807 · m / s²)
2446 · foot · m · kg / s³
Finally, to complete the progress towards MKS, we move toward the M of
2446 · foot · m · kg / s³ · (0.3048 · m) / (1 · foot)
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
(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
745.5 · (m² · kg / s³) · (Watt) / (kg · m² / s³)
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.
73's
Richard Clark, KB7QHC
Richard Clark
2003-08-20 06:01:50 UTC
Permalink
Post by k***@sympatico.ca
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.
Post by k***@sympatico.ca
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.
Post by k***@sympatico.ca
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.
Post by k***@sympatico.ca
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.

73's
Richard Clark, KB7QHC
Dr. Slick
2003-08-20 17:55:34 UTC
Permalink
Post by Richard Clark
Post by k***@sympatico.ca
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
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.
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
Richard Harrison
2003-08-21 12:51:24 UTC
Permalink
Dr. Slick wrote:
"And if the permittivity (impedance) of the material surrounding an
antenna will affect its input impedance, I think it is something to
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.

Best regards, Richard Harrison, KB5WZI
Dr. Slick
2003-08-20 05:47:34 UTC
Permalink
Post by k***@sympatico.ca
Post by 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.
Post by k***@sympatico.ca
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.
Post by k***@sympatico.ca
Post by Dr. Slick
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.
Post by k***@sympatico.ca
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.
Post by k***@sympatico.ca
Post by Dr. Slick
Post by k***@sympatico.ca
...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.
Post by k***@sympatico.ca
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
k***@sympatico.ca
2003-08-20 10:04:25 UTC
Permalink
Post by Dr. Slick
Post by k***@sympatico.ca
Actually, you've done 2*pi*radius*force work. Moving one circumference
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
2003-08-21 08:52:05 UTC
Permalink
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
2003-08-21 11:41:05 UTC
Permalink
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
Dr. Slick
2003-08-21 16:30:08 UTC
Permalink
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
Richard Clark
2003-08-21 16:48:28 UTC
Permalink
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?

73's
Richard Clark, KB7QHC
Tdonaly
2003-08-21 18:46:49 UTC
Permalink
Post by Richard Clark
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
Jim Kelley
2003-08-19 21:36:12 UTC
Permalink
Post by k***@sympatico.ca
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.
Post by k***@sympatico.ca
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. ;-)
Post by k***@sympatico.ca
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
Dr. Slick
2003-08-19 16:20:42 UTC
Permalink
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
W5DXP
2003-08-18 19:20:49 UTC
Permalink
You can still call it a transducer though - or a thigamajig.
Such a sophisticated concept deserves better. I suggest

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



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Richard Clark
2003-08-18 20:38:31 UTC
Permalink
Post by 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.
Hi Cecil,

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

73's
Richard Clark, KB7QHC
W5DXP
2003-08-19 19:29:05 UTC
Permalink
"Triactuatedmulticomplicator" or TAMC for short."
"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!
--
73, Cecil http://www.qsl.net/w5dxp



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Dr. Slick
2003-08-19 02:57:02 UTC
Permalink
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
Gene Nygaard
2003-09-24 16:45:38 UTC
Permalink
Post by Richard Clark
Post by Roy Lewallen
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
Richard Clark
2003-09-24 17:23:12 UTC
Permalink
Post by Gene Nygaard
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).

73's
Richard Clark, KB7QHC
Dave Shrader
2003-09-24 17:41:18 UTC
Permalink
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
Post by Richard Clark
Post by Gene Nygaard
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
http://physics.nist.gov/PhysRefData/contents-constants.html
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).
73's
Richard Clark, KB7QHC
Richard Clark
2003-09-24 17:58:04 UTC
Permalink
On Wed, 24 Sep 2003 17:41:18 GMT, Dave Shrader
Post by Dave Shrader
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 ;-)

73's
Richard Clark, KB7QHC
Gene Nygaard
2003-09-24 18:22:24 UTC
Permalink
Post by Richard Clark
Post by Gene Nygaard
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?
Post by Richard Clark
However, you do ask for a reference and acknowledge the NIST as a
http://physics.nist.gov/PhysRefData/contents-constants.html
There is absolutely nothing about pounds on this page. So don't be
bullshitting us.
Post by Richard Clark
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.
Post by Richard Clark
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
Richard Clark
2003-09-24 19:54:06 UTC
Permalink
Post by Gene Nygaard
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?
A balance, by implicit definition again, consists of comparing two
masses under the influence of Gravity. Given it is a bridge, in a
sense, the constant of Gravity is discarded from both sides and mass
is compared only. It is a convenience of earthly expectations (and a
defunct system of measurement) that the scale is marked in pounds.
Post by Gene Nygaard
Post by Richard Clark
However, you do ask for a reference and acknowledge the NIST as a
http://physics.nist.gov/PhysRefData/contents-constants.html
There is absolutely nothing about pounds on this page. So don't be
bullshitting us.
That is the whole point. You don't see pounds there for mass do you?
That's because pounds are not a unit of mass. They are a unit of
weight which is NOT a constant throughout the universe (nor on earth
for that matter).
Post by Gene Nygaard
Post by Richard Clark
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.
Have you tried loosing weight?
Post by Gene Nygaard
Post by Richard Clark
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.
Uh-huh. In equal measure, I couldn't "prove" that sparrows' tongues
are also "not" units of mass. Well, there are many here who's minds I
cannot change, you simply have to go to the end of that line. ;-)

73's
Richard Clark, KB7QHC
Gene Nygaard
2003-09-24 20:15:51 UTC
Permalink
Post by Richard Clark
Post by Gene Nygaard
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?
A balance, by implicit definition again, consists of comparing two
masses under the influence of Gravity. Given it is a bridge, in a
sense, the constant of Gravity is discarded from both sides and mass
is compared only. It is a convenience of earthly expectations (and a
defunct system of measurement) that the scale is marked in pounds.
The matter of convenience is in the other direction, stupid; we're
willing to substitute cheapness for accuracy in what we want to
measure on those unreliable bathroom scales. They aren't any more
accurate for measuring force than they are for measuring mass on
Earth; haven't you ever weighed yourself on your mother's scale or
somewhere else and found it differed from yours at home by several
pounds? Do you automatically assume you've gained or lost that much
weight.
Post by Richard Clark
Post by Gene Nygaard
Post by Richard Clark
However, you do ask for a reference and acknowledge the NIST as a
http://physics.nist.gov/PhysRefData/contents-constants.html
There is absolutely nothing about pounds on this page. So don't be
bullshitting us.
That is the whole point. You don't see pounds there for mass do you?
I don't see pounds as units of mass because this page just lists units
in the International System of Units.
http://w0rli.home.att.net/youare.swf

Show me something from NIST saying that pounds are not units of mass.
Or from some textbook.
Post by Richard Clark
That's because pounds are not a unit of mass. They are a unit of
weight which is NOT a constant throughout the universe (nor on earth
for that matter).
Just your say-so? That's the best you can do?
Post by Richard Clark
Post by Gene Nygaard
Post by Richard Clark
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.
Have you tried loosing weight?
To quote a sge (you know who he is) in this newsgroup:

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 ;-)
Post by Richard Clark
Post by Gene Nygaard
Post by Richard Clark
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.
Uh-huh. In equal measure, I couldn't "prove" that sparrows' tongues
are also "not" units of mass. Well, there are many here who's minds I
cannot change, you simply have to go to the end of that line. ;-)
I can, OTOH, prove that pounds are indeed units of mass.

That will prove that you are flat-out wrong in your claim that they
are not.

Just for practice, consider the troy system of weights. Unlike their
avoirdupois cousins, and unlike grams and kilograms, the troy units of
weight have never spawned units of force of the same name. They are
always units of mass; a troy ounce is exactly 31.1034768 grams, by
definition. There is not and never has been any troy pound force or
troy ounce force.

Gene Nygaard
http://ourworld.compuserve.com/homepages/Gene_Nygaard/
Richard Clark
2003-09-24 21:04:11 UTC
Permalink
Post by Gene Nygaard
Post by Richard Clark
A balance, by implicit definition again, consists of comparing two
masses under the influence of Gravity. Given it is a bridge, in a
sense, the constant of Gravity is discarded from both sides and mass
is compared only. It is a convenience of earthly expectations (and a
defunct system of measurement) that the scale is marked in pounds.
The matter of convenience is in the other direction, stupid; we're
willing to substitute cheapness for accuracy in what we want to
measure on those unreliable bathroom scales. They aren't any more
accurate for measuring force than they are for measuring mass on
Earth; haven't you ever weighed yourself on your mother's scale or
somewhere else and found it differed from yours at home by several
pounds? Do you automatically assume you've gained or lost that much
weight.
I've nowhere introduced the topic of accuracy. It has nothing to do
with your original query. Weight and mass can both be measured to
considerable accuracy. It all depends on method and standards.

A bathroom scale is not a balance. A balance has a scale (the marks
along which the balance weights are moved and the markings upon those
same weights).
Post by Gene Nygaard
Post by Richard Clark
Post by Gene Nygaard
Post by Richard Clark
However, you do ask for a reference and acknowledge the NIST as a
http://physics.nist.gov/PhysRefData/contents-constants.html
There is absolutely nothing about pounds on this page. So don't be
bullshitting us.
That is the whole point. You don't see pounds there for mass do you?
I don't see pounds as units of mass because this page just lists units
in the International System of Units.
Exactly.
Post by Gene Nygaard
Show me something from NIST saying that pounds are not units of mass.
Or from some textbook.
Post by Richard Clark
That's because pounds are not a unit of mass. They are a unit of
weight which is NOT a constant throughout the universe (nor on earth
for that matter).
Just your say-so? That's the best you can do?
I am a trained Metrologist. I have measured mass traceable to the
NIST. I have done this in four different Primary and Secondary
Standards Labs. I was the head Metrologist of two of them.
Post by Gene Nygaard
Post by Richard Clark
Have you tried loosing weight?
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 ;-)
I suppose that is an affirmative.
Post by Gene Nygaard
I can, OTOH, prove that pounds are indeed units of mass.
By a reference found at the NIST? I think you would have done that by
now if you could.
Post by Gene Nygaard
That will prove that you are flat-out wrong in your claim that they
are not.
Well, I have seen a lot of math tossed over the transom here. But if
we are to work by your own standard, cite an NIST reference.
Post by Gene Nygaard
Just for practice, consider the troy system of weights. Unlike their
avoirdupois cousins, and unlike grams and kilograms, the troy units of
weight have never spawned units of force of the same name. They are
always units of mass; a troy ounce is exactly 31.1034768 grams, by
definition. There is not and never has been any troy pound force or
troy ounce force.
Hi Gene,

Sounds like you proved a pound is not mass.

The pages I offered provide a meaningful quote:
"The 3d CGPM (1901), in a declaration intended to end the
ambiguity in popular usage concerning the word "weight," confirmed
that:
The kilogram is the unit of mass..."

Any other usage of "weight" in regard to the sensation of the action
of Gravity upon an amount of mass is outdated by more than a century
of understanding and convention.

73's
Richard Clark, KB7QHC
Gene Nygaard
2003-09-25 03:41:10 UTC
Permalink
Post by Richard Clark
Post by Gene Nygaard
Post by Richard Clark
A balance, by implicit definition again, consists of comparing two
masses under the influence of Gravity. Given it is a bridge, in a
sense, the constant of Gravity is discarded from both sides and mass
is compared only. It is a convenience of earthly expectations (and a
defunct system of measurement) that the scale is marked in pounds.
The matter of convenience is in the other direction, stupid; we're
willing to substitute cheapness for accuracy in what we want to
measure on those unreliable bathroom scales. They aren't any more
accurate for measuring force than they are for measuring mass on
Earth; haven't you ever weighed yourself on your mother's scale or
somewhere else and found it differed from yours at home by several
pounds? Do you automatically assume you've gained or lost that much
weight.
I've nowhere introduced the topic of accuracy. It has nothing to do
with your original query. Weight and mass can both be measured to
considerable accuracy. It all depends on method and standards.
And mass can be measured with much more accuracy than force can, but
that is entirely irrelevant to the point I was making.

Your claim was that the mass-measuring balances are, for a matter of
convenience, marked in units of force called pounds. I say that it is
in fact the other way around, that the cheap force-measuring spring
scales are marked in units of mass, which is indeed what we want to
measure. The kilograms used throughout the world, including most
hospitals in the U.S., for human body weight are indeed the proper SI
units for this quantity. The pounds used for this purpose are the
ones legally defined as 0.45359237 kg. Except, of course, for some
science teachers and some physics books written recently by authors so
miseducated (not uneducated, but actually mistaught) that they believe
pounds are not units of mass.

American Society for Testing and Materials, Standard for Metric
Practice, E 380-79, ASTM 1979.

3.4.1.2 Considerable confusion exists in the use
of the term weight as a quantity to mean either
force or mass. In commercial and everyday use,
the term weight nearly always means mass; thus,
when one speaks of a person's weight, the
quantity referred to is mass. . . .
Because of the dual use of the term weight as a
quantity, this term should be avoided in technical
practice except under circumstances in which its
meaning is completely clear. When the term is
used, it is important to know whether mass or
force is intended and to use SI units properly as
described in 3.4.1.1, by using kilograms for mass
or newtons for force.

This ASTM E 380 and a separate ANSI/IEEE Standard have now been
combined into a joint standard SI 10. I don't know if it says exactly
the same thing; but I am certain it doesn't say anything directly
contrary to this.

Of course, NIST also tells us the same thing. I'll get to that below.
Post by Richard Clark
A bathroom scale is not a balance. A balance has a scale (the marks
along which the balance weights are moved and the markings upon those
same weights).
Post by Gene Nygaard
Post by Richard Clark
Post by Gene Nygaard
Post by Richard Clark
However, you do ask for a reference and acknowledge the NIST as a
http://physics.nist.gov/PhysRefData/contents-constants.html
There is absolutely nothing about pounds on this page. So don't be
bullshitting us.
That is the whole point. You don't see pounds there for mass do you?
I don't see pounds as units of mass because this page just lists units
in the International System of Units.
Exactly.
So why were you offering it as evidence that pounds are not units of
mass? Do you think I'm that stupid, that you can pull the wool over
my eyes so easily? Guess again.
Post by Richard Clark
Post by Gene Nygaard
Show me something from NIST saying that pounds are not units of mass.
Or from some textbook.
Post by Richard Clark
That's because pounds are not a unit of mass. They are a unit of
weight which is NOT a constant throughout the universe (nor on earth
for that matter).
Just your say-so? That's the best you can do?
I am a trained Metrologist. I have measured mass traceable to the
NIST. I have done this in four different Primary and Secondary
Standards Labs. I was the head Metrologist of two of them.
Wow! This is even better than I dreamed of. A genuine Capital-Letter
Metrologist. Of course, it's also pretty sad, as most people will
understand if they stick with me for the rest of this message.

I'm sure that as a Metrologist, you are well aware of one particular
subset of English units, used only in calculations, which is a
coherent, gravitational foot-pound-second system in which the derived
unit of mass is a slug, equal to 1 lbf·s²/ft. One of several such
subsystems, of course.

But if you are really a capital-letter Metrologist, and an old fart on
top of it (that system with the slugs was never used in physics
textbooks before 1940, and even a couple of decades later I learned
the system I'm about to describe first, before learning the one with
slugs--and you must be at least close to my age, and a genuine expert
on weights and measures on top of it all), you'll have a damn hard
time convincing me, or anyone else, that you are also not aware of a
much older coherent foot-pound-second system of mechanical units, the
absolute fps system in which the derived unit of force is the poundal,
the force which will accelerate the base unit of mass in this system
at a rate of one foot per second squared.

Now, fill in the blank, please: The BASE UNIT OF MASS in this oldest
English system of mechanical units is the _______________. Hint: it
is the "p" in this fps system.

BTW, while the gravitational fps system of units enjoyed a brief
heyday in science in North America, outside of North America the
absolute fps system with poundals remained the system of choice for
doing calculations in English units.

You probably also know that both of these limited use, coherent
systems of units are, like SI, coherent systems of units. That means
that in neither of these do we have any pints or gallons of any sort,
not U.S. liquid, not U.S. dry, and not imperial. Nor are there any
Btu, nor horsepower, in either system. Not only that, but there are
no ounces (neither avoirdupois nor troy, nor U.S. nor imperial fluid
ounces), no tons (neither long nor short, neither force nor mass), and
no miles or inches (and thus no pounds force per square inch either).

Of course none of our ordinary measurements are made in the context of
any of these specialized systems of mechanical units which serve as
calculation aids. The fact that many of the we use are not part of
these systems is one bit of evidence of that fact. That fact that
nobody ever measures (as opposed to calculating from other
measurements) mass in slugs is another. The fact that we can
generally choose any of several different systems of units to use in
our calculations, with no change in difficulty, is still another piece
in the puzzle.

Let's look at what the English physicists William Thomson (for whom
the SI unit of temperature is named) and Peter Guthrie Tate had to say
about this way back in 1879, Treatise on natural philosophy, 1879,
reprinted as Principles of mechanics and dynamics, quoted by Jim Carr
in Apr 1998 on newsgroups alt.sci.physics, sci.engr, sci.physics.

"By taking the gravity of a constant mass for the unit
of force it makes the unit of force greater in high than
in low latitudes. In reality, standards of weight are
masses, not forces. They are employed primarily in
commerce for the purpose of measuring out a definite
quantity of matter; not an amount of matter which
shall be attracted to the earth with a given force."

<... description of merchant using spring scale to
defraud or be defrauded depending on latitude,
etc ... Jim Carr>

"It is therefore very much simpler and better to take
the imperial pound ... as the unit of mass, and to
derive from it, according to Newton's definition
above, the unit of force."


Then you might also know what "weights" means in the English versions
of the international bodies charged with keeping our international
standards:

CGPM General Conference on Weights and Measures
CIPM International Committee for Weights and Measures
BIPM International Bureau of Weights and Measures

In the introduction to their SI brochure (available at
http://www.bipm.fr), the BIPM tells us for the first half-century of
their existence, their only responsibility was keeping the standards
for length and for mass. Take a wild guess which of those two
corresponds to "weights" in these names.
Post by Richard Clark
Post by Gene Nygaard
I can, OTOH, prove that pounds are indeed units of mass.
By a reference found at the NIST? I think you would have done that by
now if you could.
Cocky little bastard, aren't you!

Just making clear that you accept the fact that doing so would prove
you wrong, before I do it.

I have already done so, of course, without referring to NIST, with
that description of the absolute fps system above.

But before we get to wandering around NIST's website, let's do a
little primary source research, and find the definition which NIST
considers controlling. For that we need to look to NIST's
predecessor, the National Bureau of Standards, and to the law of the
land.

Earlier in the 20th century, Congress had had the sense to delegate
the authority to make these definitions to the experts in the field
who know what they are doing, and had given the predecessor of NBS
this authority. This was officially implemented in 1959 by official
regulatory action by NBS, made official with its publication in the
Federal Register of 1 July 1959.

This redefinition of the pound was done in accordance with an
agreement reached among the national standards laboratories of several
of the most advanced nations in the world, not back in the Dark Ages
but in the middle of the 20th century, two years after Sputnik and the
year before the International System of Units was introduced. The old
U.S. definition, which had been a slightly different exact fraction of
a kilogram for the 66 years before then, was replaced by the new
international definition as 1 lb = 0.45359237 kg.

You can read the current U.S. law (this Federal Register Notice), as
well as a discussion of the earlier U.S. law and of the international
agreement, at one of these web sites (same document both places):
http://www.ngs.noaa.gov/PUBS_LIB/FedRegister/FRdoc59-5442.pdf
http://gssp.wva.net/html.common/refine.pdf

Of course, the same definition is was also adopted in Canada, in the
U.K., in South Africa, in Australia, and in New Zealand, the other
parties to this international agreement. It is also used all around
the world, and was adopted by statute or regulation in some other
countries not a party to the original agreement, such as Ireland.

Now, let's get to the NIST web site. You obviously didn't follow the
links you provided to get to the right place. Start with the one you
recommended
http://physics.nist.gov/cuu/Units/units.html
click on Return to Units Home Page which takes you back up one level
to
http://physics.nist.gov/cuu/Units/index.html
click on Bibliography: Online publications and citations to go to
http://physics.nist.gov/cuu/Units/bibliography.html
Go down to NIST Special Publication 811 and first of all, order
yourself a free printed copy of this document. You need it. Then
look at it in the online version, either in .html or in .pdf,
whichever you prefer. I'll use the html version to give some links to
specific parts of it below; I also have the printed version, and I
have the .pdf version right on my computer.

I already know what you think of NIST. In another message, you told
us that "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. "

This is NIST's official _Guide for the Use of the International System
of Units (SI)_, by Dr. Barry N. Taylor, reviewed and approved by both
the director of NIST and by his boss, the Secretary of Commerce. It
is cited as an authority not only by many of the national standards
agencies around the world, but also by the BIPM itself.

First, a word about Dr. Barry N. Taylor. He is not only a
professional metrologist with a Ph.D. in physics, but he has also
served on both the Consultative Committe on Units which advises the
CGPM (still on that, I think), and on the SUNAMCO Commission
(Commission on Symbols, Units, Nomenclature, Atomic Masses and
Fundamental Constants) of the International Union of Pure and Applied
Physics (IUPAP).

In other words, for the benefit of anyone else reading this, if
Richard Clark is a "Metrologist" then Dr. Barry Taylor must be a
"METROLOGIST" because even capitalizing all the letters won't
adequately show the difference between the two, especially when it
comes to expertise in the particular subfield related to teh
definitions of units of measure.

First, before we get to the pounds, let's digress a little bit and
finish up that loose end I left above.
http://physics.nist.gov/Pubs/SP811/sec08.html

Thus the SI unit of the quantity weight used in this
sense is the kilogram (kg) and the verb "to weigh" means
"to determine the mass of" or "to have a mass of".

Examples: the child's weight is 23 kg

The same is true for pounds, of course. Units of mass in this
context, as the term is used in physiology and medicine, and in
sports--the reasons we normally weigh ourselves.

There's more to the explanation in section 8.3, including a good
discussion of the force definition of weight often used in physics and
engineering. This section concludes with the excellent advice that

whenever the word 'weight' is used, it should be
made clear which meaning is intended.

Now, let's go to the extensive list of conversion factors found in
Appendix B to SP 811.
http://physics.nist.gov/Pubs/SP811/appenB8.html#P

To convert from to Multiply by

pound (avoirdupois) (lb) 23 kilogram (kg) 4.535 924 E-01
pound (troy or apothecary) (lb) kilogram (kg) 3.732 417 E-01

[The 23 is a reference to a footnote in the printed and pdf versions,
a note on a separate page in html]
http://physics.nist.gov/Pubs/SP811/footnotes.html#f23

23 The exact conversion factor is 4.535 923 7 E-01. All units
in Sec. B.8 and Sec. B.9 that contain the pound refer to
the avoirdupois pound.

This unit, of course, is not defined by this publication. This is
just the legal definition made by NBS in 1959.

Also, take a look at another section of American Society for Testing
and Materials, Standard for Metric Practice, E 380-79, ASTM 1979.

3.4.1.4 The use of the same name for units of force
and mass causes confusion. When the non-SI units
are used, a distinction should be made between
force and mass, for example, lbf to denote force in
gravimetric engineering units and lb for mass.

As you can see above, this sensible rule is also followed by NIST. It
is also followed by NPL, the U.K. national standards laboratory. It
is the older unit, the one more often used, and the one more likely to
be used by those who care least about the distinction, which gets to
use the original, unadorned symbol "lb"; it is the recent
bastardization, the less often used unit, and the one more likely to
be used by those who care most about the distinction, that must be
distinguished by using a different symbol, "lbf" instead.
Post by Richard Clark
Post by Gene Nygaard
That will prove that you are flat-out wrong in your claim that they
are not.
Well, I have seen a lot of math tossed over the transom here. But if
we are to work by your own standard, cite an NIST reference.
Post by Gene Nygaard
Just for practice, consider the troy system of weights. Unlike their
avoirdupois cousins, and unlike grams and kilograms, the troy units of
weight have never spawned units of force of the same name. They are
always units of mass; a troy ounce is exactly 31.1034768 grams, by
definition. There is not and never has been any troy pound force or
troy ounce force.
Hi Gene,
Sounds like you proved a pound is not mass.
No. You just proved that you are hopelessly ignorant when you get
outside your fields and start discussing things such as linguistics,
history, or the law.
Post by Richard Clark
"The 3d CGPM (1901), in a declaration intended to end the
ambiguity in popular usage concerning the word "weight," confirmed
The kilogram is the unit of mass..."
One of the most confusing and impossible to understand resolutions any
political body has ever passed. You will note that NIST places no
emphasis on this whatsoever.

Yes, even then there were evidently enough scientists so utterly
confused as to think that the standards they were keeping were
standards of force rather than the standards of mass which they always
had been. As were the old standards for pounds, naturally. Of
course, by that time, we in the United States has already abandoned
our independent standards for pounds, and we already defined them as
an exact fraction of a kilogram. So where does that lead you?

But of course, you are making a big mistake you think that this
particular resolution meant that we couldn't use kilograms force. In
fact, it was just the opposite--this very same resolution endorsed the
use of grams force and kilograms force by adopting a standard
acceleration of gravity, which is not a concept of physics but rather
of metrology, something which serves no purpose other than defining
units of force in terms of units of mass. Kilograms force had never
been well-defined units before then. Neither had pounds force, of
course--and what's more, even today pounds force don't have an
official definition. The de facto standard, never officially adopted
by any national or international standards agency, nor by any
professional organization, is probably to use the same standard
acceleration of gravity which is official for defining grams force and
pounds force, namely 980.665 cm/s² in the units used in that 1901
resolution, before mks systems had come into use, at a time when
neither slugs nor newtons had ever been used for the units they are
now.

Now go back to that Bibliography page on NIST and download NIST
Special Publication 330, which includes the unofficial English version
of this resolution, or at least the salient parts of it.
http://physics.nist.gov/cuu/Units/bibliography.html

Then pay special attention to the footnote added by NIST, found on
page 17 in the document's own pagination (I don't know the page number
in the pdf format):

[dagger] USA Editor's note: In the USA, ambiguity exists in
the use of the term weight as a quantity to mean either
force or mass. In science and technology this declaration
[CGPM (1901)] is usually followed, with the newton the
SI unit of force and thus weight. In commercial and
everyday use, weight is often used in the sense of mass
for which the SI unit is the kilogram.
Post by Richard Clark
Any other usage of "weight" in regard to the sensation of the action
of Gravity upon an amount of mass is outdated by more than a century
of understanding and convention.
You said weight is a force. This resolution clearly said that weight
is not a force, but merely something "in the nature of a force,"
whatever the hell that is supposed to mean.

But fortunately, in any case, nobody was ever damn fool enough to give
the 1901 CGPM any say-so on what "weight" means for the "net weight"
of my bag of sugar, or for the troy weight of a bar of gold or
platinum. That's outside their authority.

That resolution isn't seriously offered as proof of any change in
meaning of the word. In fact, I doubt that the 1901 CGPM ever
intended to change the meaning of the word--they wrongly thought that
they were merely stating existing definitions.

"Outdated by more than a century of understanding and convention"?
Nonsense. Go back and read that section 8.3 of NIST Special
Publication 811 again. That's 1995--hardly a century ago.

Need more. Here's 1989, in the still-effective official National
Standard of Canada, CAN/CSA-Z234.1-89 Canadian Metric
Practice Guide:

5.7.3 Considerable confusion exists in the use of
the term "weight." In commercial and everyday use,
the term "weight" nearly always means mass. In
science and technology, "weight" has primarily
meant a force due to gravity. In scientific and
technical work, the term "weight" should be
replaced by the term "mass" or "force," depending
on the application.

5.7.4 The use of the verb "to weigh" meaning "to
determine the mass of," e.g., "I weighed this
object and determined its mass to be 5 kg,"
is correct.

Note that "nearly always" is much stronger than "primarily"; they even
got that part right. Note further the difference usage for the noun
forms in 5.7.3 and the verb forms in 5.7.4; for the former, the
meaning is context-specific, but for the latter that definition is
unqualifiedly called "correct" (which does not, of course, say
anything one way or the other about the use of the verb to mean to
determine the force due to gravity, which is also correct).

Need more. Here's 2003, on the web pages on the National Physical
Laboratory (NPL), the official national standards agency of the United
Kingdom of Great Britain and Northern Ireland:

NPL FAQ
http://www.npl.co.uk/force/faqs/forcemassdiffs.html

Weight
In the trading of goods, weight is taken to mean the
same as mass, and is measured in kilograms. Scientifically
however, it is normal to state that the weight of a body is
the gravitational force acting on it and hence it should be
measured in newtons, and this force depends on the local
acceleration due to gravity. To add to the confusion, a
weight (or weightpiece) is a calibrated mass normally
made from a dense metal, and weighing is generally
defined as a process for determining the mass of an
object.

So, unfortunately, weight has three meanings and care
should always be taken to appreciate which one is
meant in a particular context.

Note that they are talking about DIFFERENT MEANINGS of the word
"weight." Just as NIST does. Just as ASTM does. Just as the
Canadian Standard for Metric Practice does. Just as any good
dictionary does.

That, of course, is how the word weight entered the English language
over 1000 years ago, meaning the quantity measured with a balance. A
quantity which you yourself explained so lucidly to be mass, not the
force due to gravity. Of course, when those tribesmen in what is now
England were looking for a word to use to measure how much stuff they
have, when they buy and sell goods, they didn't make any mistake when
they invented this word "weight" for that purpose, did they? They
couldn't have used "mass" for this quantity instead, unless they had
happened to choose those phonemes for the word they invented. Mass
didn't have this meaning until more than 750 years later, when some
obscure translator translated Newton's major work into English after
Newton's death. Or do you think it was a mistake that these heathens
didn't figure out the God-given word they were supposed to invent for
this purpose?

There is no error when we use the very same word, with the very same
meaning, for the very same purposes today. We have a prior claim to
this word by 3/4 of a millennium over the physicists who recently
borrowed it and often use it with a somewhat different meaning.
Sometimes we borrow the physicists meaning, sometimes we use the
original meaning. But like all the experts tell you, because of these
ambiguities, you should just avoid using the term "weight" in a
technical context--and if you do use it, make damn sure your meaning
is clear.

Gene Nygaard
http://ourworld.compuserve.com/homepages/Gene_Nygaard/
Gentlemen of the jury, Chicolini here may look like an idiot,
and sound like an idiot, but don't let that fool you: He
really is an idiot.
Groucho Marx
Richard Clark
2003-09-25 07:14:09 UTC
Permalink
Post by Gene Nygaard
Do you think I'm that stupid, that you can pull the wool over
my eyes so easily?
Hi Gene,

As I pointed out earlier, your feelings belong at the end of the line
with the rest whose minds I cannot change.

73's
Richard Clark, KB7QHC
Dave Shrader
2003-09-25 01:16:33 UTC
Permalink
Gene Nygaard wrote:

[SNIP]
Post by Gene Nygaard
Apparently you are claiming that pounds are not units of mass.
Where did you learn that?
Well, I learned that a Pound is a unit of Force.
Well, I learned that a Slug [pound mass] is Pound*acceleration.
Well, I learned that mass is pound*sec^2/foot.

Where did I learn this? What's my source? Physics 101, University
Physics, Sears and Zemansky, Addison-Wesley Publishing, 1956, Chapter 6,
page 94.

I hope tou don't need another reference?

Now, what's your real problem? What are you trying to say?

Dave, W1MCE
Post by Gene Nygaard
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
2003-09-25 01:27:41 UTC
Permalink
On Thu, 25 Sep 2003 01:16:33 GMT, Dave Shrader
Post by Dave Shrader
[SNIP]
Post by Gene Nygaard
Apparently you are claiming that pounds are not units of mass.
Where did you learn that?
Well, I learned that a Pound is a unit of Force.
Well, I learned that a Slug [pound mass] is Pound*acceleration.
Well, I learned that mass is pound*sec^2/foot.
Where did I learn this? What's my source? Physics 101, University
Physics, Sears and Zemansky, Addison-Wesley Publishing, 1956, Chapter 6,
page 94.
I hope tou don't need another reference?
Now, what's your real problem? What are you trying to say?
Can you quote it to me, specifically where it says that pounds are not
units of mass? I'll bet you just misunderstood what it said. I have
the 1970 edition of Sears and Zemansky myself, so I'm betting that if
anything, what it actually says is clearer in that older edition than
it is in the 1970 edition.

Pounds force do exist, of course. What I'm asking you to show me is
not that, but rather that pounds are not units of mass.

Sears and Zemansky didn't lie about this in 1956. They might have
been dishonest and deceptive about it, not concerned enough about the
possibility that fools like you would misinterpret what they said or
actually encouraging such misinterpretation. But they didn't lie
about it. Some textbooks today might actually lie about it (or,
alternatively, their authors are too poorly educated to know any
better--take your choice).

Gene Nygaard
Post by Dave Shrader
Dave, W1MCE
Post by Gene Nygaard
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
http://ourworld.compuserve.com/homepages/Gene_Nygaard/
Dave Shrader
2003-09-25 12:13:47 UTC
Permalink
Gene, thanks for the compliment in calling the Program Chief Engineer
of the USAF MX [Peacekeeper] Re-Entry System/Re-Entry Vehicle a fool.

It says a lot about you. I forgive you.

Dave, W1MCE
+ + +
not concerned enough about the possibility that fools like you
Gene Nygaard
2003-09-25 12:47:53 UTC
Permalink
On Thu, 25 Sep 2003 12:13:47 GMT, Dave Shrader
Post by Dave Shrader
Gene, thanks for the compliment in calling the Program Chief Engineer
of the USAF MX [Peacekeeper] Re-Entry System/Re-Entry Vehicle a fool.
It says a lot about you. I forgive you.
Gee, if I'd known you were so important, I'd really have taken you to
task for being too damn stupid to understand what you read in Sears
and Zemansky!

Gene Nygaard
http://ourworld.compuserve.com/homepages/Gene_Nygaard/
Gene Nygaard
2003-09-25 15:20:45 UTC
Permalink
On Thu, 25 Sep 2003 12:13:47 GMT, Dave Shrader
Post by Dave Shrader
Gene, thanks for the compliment in calling the Program Chief Engineer
of the USAF MX [Peacekeeper] Re-Entry System/Re-Entry Vehicle a fool.
It says a lot about you. I forgive you.
Dave, W1MCE
+ + +
not concerned enough about the possibility that fools like you
Since you aren't honest enough to tell us exactly what Sears and
Zemansky said in 1956, I'll tell everyone what they said in 1970. If
there are any significant differences, feel free to point them out.
This thing is, I know that Sears and Zemansky weren't going to lie
about this, because they grew up using poundals, which are by
definition the force which will accelerate a MASS of 1 lb at a rate of
1 ft/s².

Francis Weston Sears and Mark W. Zemansky, University Physics,
Addison-Wesley, 4th ed., 1970.

[page 3]

1 pound mass = 1 lbm = 0.45359237 kg

[The actual number will, of course, be different in 1956, because the
U.S. didn't adopt this definition until 1959 (it had been in use in
Canada since 1953, six years before the international
redefinition).--GAN]

[page 4]

We select as a standard body the standard pound,
defined in section 1-2 as a certain fraction
(approximately 0.454) of a standard kilogram.

[page 59]

In setting up the mks and cgs systems, we first selected
units of mass and acceleration, and defined the unit of
force in terms of these. In the British engineering system,
we first select a unit of force (1 lb) and a unit of
acceleration (1 ft s^-2) and then define the unit of mass as
the mass of a body whose acceleration is 1 ft s^-2 when
the resultant force on the body is 1 lb.

<end quote>

Now, Sears and Zemansky might be incompetent for not allowing for the
fact that there are going to be people out there who are too blamed
stupid to understand that that adjectival phrase "British engineering"
has some meaning, and that it identifies one particular limited subset
of the British units. It's perhaps even understandable, because that
fact would be quite clear to anyone who, like them, had grown up using
poundals in a "British absolute" system of units.

However, that doesn't change the fact that you are in fact one of the
people who are that stupid.
--
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
Jim Kelley
2003-09-25 18:51:47 UTC
Permalink
"Fundamentals of Physics", Haliday and Resnick, Second Edition, 1981

Appendix F, Conversion Factors

Mass

"Quantities in the colored areas [ounce, pound, ton] are not mass units
but are often used as such. When we write, for example 1 kg "=" 2.205
lb this means that a kilogram is a _mass_ that _weighs_ 2.205 pounds
under standard condition of gravity (g = 9.80665 m/s^2)."

The units dyne, Newton, pound, and poundal are listed elsewhere in
Appendix F as units of force.

73, AC6XG
Post by Gene Nygaard
On Thu, 25 Sep 2003 12:13:47 GMT, Dave Shrader
Post by Dave Shrader
Gene, thanks for the compliment in calling the Program Chief Engineer
of the USAF MX [Peacekeeper] Re-Entry System/Re-Entry Vehicle a fool.
It says a lot about you. I forgive you.
Dave, W1MCE
+ + +
not concerned enough about the possibility that fools like you
Since you aren't honest enough to tell us exactly what Sears and
Zemansky said in 1956, I'll tell everyone what they said in 1970. If
there are any significant differences, feel free to point them out.
This thing is, I know that Sears and Zemansky weren't going to lie
about this, because they grew up using poundals, which are by
definition the force which will accelerate a MASS of 1 lb at a rate of
1 ft/s².
Francis Weston Sears and Mark W. Zemansky, University Physics,
Addison-Wesley, 4th ed., 1970.
[page 3]
1 pound mass = 1 lbm = 0.45359237 kg
[The actual number will, of course, be different in 1956, because the
U.S. didn't adopt this definition until 1959 (it had been in use in
Canada since 1953, six years before the international
redefinition).--GAN]
[page 4]
We select as a standard body the standard pound,
defined in section 1-2 as a certain fraction
(approximately 0.454) of a standard kilogram.
[page 59]
In setting up the mks and cgs systems, we first selected
units of mass and acceleration, and defined the unit of
force in terms of these. In the British engineering system,
we first select a unit of force (1 lb) and a unit of
acceleration (1 ft s^-2) and then define the unit of mass as
the mass of a body whose acceleration is 1 ft s^-2 when
the resultant force on the body is 1 lb.
<end quote>
Now, Sears and Zemansky might be incompetent for not allowing for the
fact that there are going to be people out there who are too blamed
stupid to understand that that adjectival phrase "British engineering"
has some meaning, and that it identifies one particular limited subset
of the British units. It's perhaps even understandable, because that
fact would be quite clear to anyone who, like them, had grown up using
poundals in a "British absolute" system of units.
However, that doesn't change the fact that you are in fact one of the
people who are that stupid.
--
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
Cecil Moore
2003-09-25 19:09:04 UTC
Permalink
Post by Jim Kelley
"Quantities in the colored areas [ounce, pound, ton] are not mass units
but are often used as such.
What is the mass of a banana slug in slugs?
--
73, Cecil, W5DXP
Jim Kelley
2003-09-25 19:23:34 UTC
Permalink
Post by Cecil Moore
Post by Jim Kelley
"Quantities in the colored areas [ounce, pound, ton] are not mass units
but are often used as such.
What is the mass of a banana slug in slugs?
Ask somebody at UC Santa Cruz. ;-)

73, ac6xg
Dave Shrader
2003-09-25 20:06:37 UTC
Permalink
You are equating pound and POUNDAL ['pound mass']. They are two
different things.

----------------------------------------

Sears and Zemansky, 1956, Table 5-1, page 77

Systems of units Force Mass Acceleration
Engineering pound Slug ft/sec^2
mks newton kilogram m/sec^2
cgs dyne gram cm/sec^2
----------------------------------------

"One standard pound, by definition, is a body of mass 0.4535924277 kg."

"Since the weight of a body is a force, it must be expressed in units of
force. Thus in the engineering system weight is expressed in POUNDS; in
the mks system, in Newtons; and in the cgs system, in dynes."

Unless you disagree with Newton's Second Law, F=ma, Force [pounds] and
mass [slugs] are related by acceleration [of gravity, for example].

So, my weight [240 pounds] = my mass [7.45 slugs]*[gravity of 32.2
ft/sec^2].
-----------------------------------------
If you want to argue, go ahead. I cited a source as you asked. Now you
choose to disagree with that source.

My final comment: Does a newton[force] = a kilogram[mass]??
--------------

Conclusion:

Force = pounds, or newtons, or dynes.
Mass = Slug, or kilogram, or gram
Acceleration = ft/sec^2, or m/sec^2, or cm/sec^2
----------------------------------------
Don't be so everbearing! It does not become you or enhance you statements.
Gene Nygaard
2003-09-25 15:38:45 UTC
Permalink
On Thu, 25 Sep 2003 12:13:47 GMT, Dave Shrader
Post by Dave Shrader
Gene, thanks for the compliment in calling the Program Chief Engineer
of the USAF MX [Peacekeeper] Re-Entry System/Re-Entry Vehicle a fool.
Is that as good as being chief of the Mars Climate Orbiter program as
evidence of competence in the use of units of measure?

Gene Nygaard
http://ourworld.compuserve.com/homepages/Gene_Nygaard/
k***@sympatico.ca
2003-09-25 10:45:29 UTC
Permalink
Post by Gene Nygaard
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
Nice web page you have on the subject, but I suspect it is not
quite so cut and dried as you make out.

I have a strong recollection (from many years ago) of being
taught that pounds where force. Going to google with 'pound
mass force' yields some modern university teaching material
which says the same. My ancient thermo text uses lbf and lbm
throughout to eliminate confusion.

There seems to be little doubt that today the pound is
defined in terms of the kilogram so is clearly a unit of
mass.

But usage of the pound seems to be less consistent. Consider
pounds per square inch or foot-pounds; in each of these
the pound is a unit of force.

I expect the definition of pound will be argued for some
years to come.

Let's just all go metric. The only really confusing measure
there seems to be the definition of the litre.

...Keith
Gene Nygaard
2003-09-25 12:36:32 UTC
Permalink
Post by k***@sympatico.ca
Post by Gene Nygaard
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
Nice web page you have on the subject, but I suspect it is not
quite so cut and dried as you make out.
It is. I never claimed that pounds force aren't used also.
Post by k***@sympatico.ca
I have a strong recollection (from many years ago) of being
taught that pounds where force. Going to google with 'pound
mass force' yields some modern university teaching material
which says the same. My ancient thermo text uses lbf and lbm
throughout to eliminate confusion.
But that isn't what our resident Capital-M Metrologist was telling us.
He claimed flatout that pounds are not units of mass.

Furthermore, he also set the ground rules that I could prove him wrong
from the NIST web pages. Yet even though I have indeed done not only
that, and with a dozen other reasons and citations as well, he remains
recalcitrant in his erroneous ways.

Richard Clark is a big windbag, of no substance whatsoever. But much
worse than that is the fact that he has absolutely no integrity.

Yes, some do use lbm and lbf. Or some other variation. I remember
one textbook using p for pounds force and lb for pounds mass.

Here's the way the British physicist who named the slug put in, in the
little treatise in which he introduced this unit to the world:

In the interests of clear teaching, the convention
(which I am glad to see has been adopted in America)
has been adhered to throughout, of using the word
‘pound' when a force is meant, and ‘lb.' when a mass
is meant, and I have ventured to give the name of a
‘slug to the British Engineer's Unit of Mass, i.e. to the
mass in which an acceleration of one foot-per-sec.-
per-sec. is produced by a force of one pound.

It's easy to see why that convention didn't go very far. People would
naturally like to have both a spelled out word and a symbol for both
pounds. He also says, another source showing that Richard Clark is an
incompetent metrologist (and if is also a professional engineer,
guilty of malpractice by his own words because he doesn't follow
NIST's definitions):

British Absolute Unit of Torque. Since in the British
absolute system, in which the lb. is chosen as the unit
of mass, the foot as unit of length, and the second as
unit of time, the unit of force is the poundal, it is
reasonable and is agreed that the British absolute
unit of torque shall be that of a poundal acting at a
distance of 1 foot, or (what is the same thing, as regards
turning) a couple of which the force is one poundal and
the arm one foot. This we shall call a poundal-foot,
thereby distinguishing it from the foot-poundal, which
is the British absolute unit of work.

A.M. Worthington, Dynamics of Rotation: An Elementary Introduction of
Rigid Dynamics, London, New York, Bombay, Calcutta, and Madras:
Longmans, Green, and Co., 1920

As I pointed out to Richard Clark, however, according to the modern
rules, it is the pound force which must be distinguished from pounds
in their original and still most common meaning as units of mass. The
ones using pounds mass (the unit much more often used by those who
care least about the distinction--witness the thousands of items in
every grocery store and hardware store in America measured in pounds)
are the ones who can get away with using the original "lb" symbol.
Post by k***@sympatico.ca
There seems to be little doubt that today the pound is
defined in terms of the kilogram so is clearly a unit of
mass.
I can't understand how so many science teachers, and even textbook
authors in the past decade or so, can remain oblivious to this simple
fact.
Post by k***@sympatico.ca
But usage of the pound seems to be less consistent. Consider
pounds per square inch or foot-pounds; in each of these
the pound is a unit of force.
Yes, and for all those items in our stores, the pound is a unit of
mass. For density in pounds per cubic foot or pounds per gallon or
pounds per cubic yard (civil engineers) or pounds per cubic inch
(mechanical engineers), pounds are units of mass. For defining a
British thermal unit, how much water? Certainly not an amount that
exerts a certain amount of force. By the same token, the pounds in
the denominator are units of mass if you say the latent heat of fusion
of water is 80 Btu/lb, or if you express specific heat capacity in
units of Btu/(lb·°F).

Of course, we also have kilograms as units of force when you see the
thrust of jet or rocket engines expressed in kilograms, or pressure
gauges in kg/cm², or torque wrenches in "meter kilograms" rather than
newton meters (still readily available; I have one). The only
difference is that these uses are decreasing. That's because the
metric system is still fully supported and updated, and kilograms
force are not part of its modern version, the International System of
Units (SI). The English units, OTOH, are like obsolete versions of
software--still often usable for the purposes the user wants, but
eventually you may have to upgrade if you want to communicate with the
rest of the world. Nobody is ever going to bother to tell us to stop
using pounds force, without telling us to stop using pounds as units
of mass as well.
Post by k***@sympatico.ca
I expect the definition of pound will be argued for some
years to come.
Let's just all go metric. The only really confusing measure
there seems to be the definition of the litre.
The litre, of course, is not part of SI. While its definition was
confusing around the time I first learned this, when it was different
from a cubic decimeter, that changed in 1964 when the CGPM redefined
it, restoring the definition as exactly 1 dm^3, and said that under
that definition it would be acceptable for use with SI, something that
hadn't been true for the first four years of the existence of SI.

No. You have the mole, which shares its name with the pound mole and
the kilogram mole, both still used. Guess nobody has the political
will to rename it the loschmitt.

For other non-SI units, of course, the calorie is the absolute worse,
with several different flavors in each of two major groupings,
differing from each other by a factor of a thousand.

Gene Nygaard
http://ourworld.compuserve.com/homepages/Gene_Nygaard/
Richard Clark
2003-09-25 15:58:11 UTC
Permalink
Post by k***@sympatico.ca
Let's just all go metric. The only really confusing measure
there seems to be the definition of the litre.
...Keith
Hi Keith,

You mean liter? ;-)

73's
Richard Clark, KB7QHC
Gene Nygaard
2003-09-25 16:08:24 UTC
Permalink
Post by Richard Clark
Post by k***@sympatico.ca
Let's just all go metric. The only really confusing measure
there seems to be the definition of the litre.
...Keith
Hi Keith,
You mean liter? ;-)
Up there in the Great White North, they use those dinky little
"litres" where it takes 4.54609 of them to make a gallon, rather than
the man-sized liters we have, which only take 3.785411784 to make a
gallon. ;-)

Unless, of course, you are talking about blueberries, where we use an
inbetween liter where it takes 4.40488377086 liters to make a gallon
(which we actually don't use much under that name any more, though we
do still use its quart and pint subdivisions).

Gene Nygaard
Jim Kelley
2003-09-25 19:02:12 UTC
Permalink
Post by Gene Nygaard
Up there in the Great White North, they use those dinky little
"litres" where it takes 4.54609 of them to make a gallon, rather than
the man-sized liters we have, which only take 3.785411784 to make a
gallon. ;-)
I suspect it's not the litre which is different, but the gallon which is
different. The British Imperial Gallon occupies 277.4 in^3, while the
gallon you're thinking of occupies 231 in^3.

What's your opinion of converting US speedometers from miles/hr to
furlongs/fortnight?

73, AC6XG
k***@sympatico.ca
2003-09-25 23:11:42 UTC
Permalink
Post by Gene Nygaard
Post by Richard Clark
Post by k***@sympatico.ca
Let's just all go metric. The only really confusing measure
there seems to be the definition of the litre.
...Keith
Hi Keith,
You mean liter? ;-)
It has to be litre so that it can rhyme with metre.
Post by Gene Nygaard
Up there in the Great White North, they use those dinky little
"litres" where it takes 4.54609 of them to make a gallon, rather than
the man-sized liters we have, which only take 3.785411784 to make a
gallon. ;-)
Unless, of course, you are talking about blueberries, where we use an
inbetween liter where it takes 4.40488377086 liters to make a gallon
(which we actually don't use much under that name any more, though we
do still use its quart and pint subdivisions).
We also have the Texas sized foot of 12.789 inches (legal for
surveying only in Quebec, they say).

But it seems that in the great country to the south there are
also two definitions for the foot: 0.3048 meter and
1200/3937 meter. When I buy a tape measure made in the U.S.A.
am I getting long feet or short feet?

...Keith
Gene Nygaard
2003-09-26 00:30:01 UTC
Permalink
Post by k***@sympatico.ca
Post by Gene Nygaard
Post by Richard Clark
Post by k***@sympatico.ca
Let's just all go metric. The only really confusing measure
there seems to be the definition of the litre.
...Keith
Hi Keith,
You mean liter? ;-)
It has to be litre so that it can rhyme with metre.
Post by Gene Nygaard
Up there in the Great White North, they use those dinky little
"litres" where it takes 4.54609 of them to make a gallon, rather than
the man-sized liters we have, which only take 3.785411784 to make a
gallon. ;-)
Unless, of course, you are talking about blueberries, where we use an
inbetween liter where it takes 4.40488377086 liters to make a gallon
(which we actually don't use much under that name any more, though we
do still use its quart and pint subdivisions).
We also have the Texas sized foot of 12.789 inches (legal for
surveying only in Quebec, they say).
Interestingly enough, Thomas Jefferson used Isaac Newton's
measurements of the length of a seconds pendulum at various latitudes
in terms of these feet, the royal foot of Paris, to calculate how long
his foot would be in terms of the English feet then in use, when he
proposed a decimal system based on the foot in 1790, before the metric
system had been invented. In Jefferson's system, a metre would have
been a cubic inch (0.001 cubic foot), and a metre of cool water would
have a mass of 1 ounce (0.1 pound), and an ounce of 11/12 silver would
have been a dollar.
Plan for establishing uniformity in the Coinage, Weights, and Measures
of the United States.
http://ourworld.compuserve.com/homepages/Gene_Nygaard/t_jeff.htm#from2

Some surveys in the United States (Lousiana) were originally done in
these feet also, with lengths and areas in arpents. Some land grants
in southwestern states were in terms of varas of various sizes, with
areas expressed in labors and leagues.
Post by k***@sympatico.ca
But it seems that in the great country to the south there are
also two definitions for the foot: 0.3048 meter and
1200/3937 meter. When I buy a tape measure made in the U.S.A.
am I getting long feet or short feet?
The short ones, of course. The same document I cited before, the
Federal Register notice which is the U.S. law redefining the yard as
0.9144 m and the pound as 0.45359237 kg spells out the limited
surveying purposes for which the old definition would continue to be
used.
http://www.ngs.noaa.gov/PUBS_LIB/FedRegister/FRdoc59-5442.pdf
http://gssp.wva.net/html.common/refine.pdf

Seriously, you are buying a lot better quality tape measures than I
have ever used, if you expect them to be accurate to that 2 parts per
million difference. Are any tape measures that good? You will, of
course, have to be making temperature corrections as well.

Gene Nygaard
http://ourworld.compuserve.com/homepages/Gene_Nygaard/

Dr. Slick
2003-08-17 18:06:07 UTC
Permalink
Post by Roy Lewallen
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
Gene Nygaard
2003-09-24 16:46:46 UTC
Permalink
Post by Roy Lewallen
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/
Dick Carroll;
2003-08-08 16:39:25 UTC
Permalink
Post by Reg Edwards
An antenna's radiating efficiency has nothing whatever to do with the
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.
The reason they might do that would be concern for the antenna *system*
efficiency and not
just that of the antenna.

Dick
Richard Harrison
2003-08-19 19:19:45 UTC
Permalink
Richard Clark wrote:
"So how do the "ohms" of free space differ from the "ohms" of a quarter
wave monopole`s radiation resistance?"

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

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

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

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

Best regards, Richard Harrison, KB5WZI
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