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

Such a sophisticated concept deserves better. I suggest

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



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

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

Such a sophisticated concept deserves better. I suggest

"Triactuatedmulticomplicator" or TAMC for short.

Hi Cecil,

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

73's
Richard Clark, KB7QHC

Richard Clark wrote in message . ..

Zc = 376.730 · ohms

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



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


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


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

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


Slick

Richard Clark wrote in message . ..

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

73's
Richard Clark, KB7QHC


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


Slick

"Dr. Slick" wrote:

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

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

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

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

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

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

....Keith

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

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

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

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


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

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

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

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

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

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

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

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

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

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

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

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


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

...Keith

Hi Keith,

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

From page 3 (yes, pretty up front):

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

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

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

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

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

73's
Richard Clark, KB7QHC

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

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

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

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


Not really. Look at this:

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

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

interesting that you bring this up.



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



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

Look he

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

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

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

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

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

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




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

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

...Keith


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


Slick

"Dr. Slick" wrote:

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

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

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

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


Not really. Look at this:

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

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

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

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

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

interesting that you bring this up.


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


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

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

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

Look he

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

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

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

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

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

But the crux is
that angles are dimensionless:

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

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


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

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

...Keith

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

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

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

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

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

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

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

....Keith

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

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

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

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

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

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:

Cecil, W5DXP wrote:
"Triactuatedmulticomplicator" or TAMC for short."

I have a suggested update:
"Triactuatedmultiuncomplicator", or TAMU for short.

Richard, when I was there in the 50's, it was TAMC.
My '59 graduation ring says "A&M College of Texas".
Trivia note: At that time, Texas University was a
branch of the Texas A&M system. Gig 'Um!
--
73, Cecil http://www.qsl.net/w5dxp



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

Torque is the same as work. Work is energy conversion.

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

Yes. Sorta like knowing when it's legal to carry the one. ;-)

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

Meters cancel meters.

I recently saw somewhere on the web where a given light intensity was
converted to the equivalent free space voltage and current. I think it
was a response to Dr. Slicks inquiry over on sci.physics.electromag.

.....yes, here it is.

http://www.flashrock.com/upload/photong/photong.html

73 de ac6xg

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.

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

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

. . .

On Tue, 19 Aug 2003 14:47:58 -0700, Roy Lewallen
wrote:

I find this most interesting. As a P.E. licensed by the state of Oregon
(since 1981), I'm aware that I'm subject to state laws governing the
code of conduct of Professional Engineers, and all other applicable
state laws. I didn't realize that I had legal obligations to NIST, or
that any other federal agency has requirements for P.E.s of all states.
Would you please provide some reference where I can further research
this obligation and the rules it has imposed that I'm legally required
to comply with?

Roy Lewallen, W7EL, P.E.


Hi Roy,

I am wholly unaware of the full scope of your business and contracts
and I have no interest, nor do I think you would volunteer that
information. I cannot recall a single instance of your relating any
experience of yours that revolved around the matters I have discussed,
nor any matters that were professional beyond your product. I cannot
imagine that your product enters into matters of traceability or
authority when I have seen your disclosures that explicitly remove
yourself from liability:
Legal Disclaimer

The licensee ("Licensee" or "User") acknowledges that the reliability
of any and all results produced by this software are not precise and
are subject to significant levels of variability.
....
LICENSOR HEREBY DISCLAIMS ANY AND ALL WARRANTIES OF
MERCHANTABILITY OR FITNESS FOR A PARTICULAR PURPOSE.

73's
Richard Clark, KB7QHC

On Tue, 19 Aug 2003 14:35:55 -0400, wrote:

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

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

Richard Clark wrote:

On Tue, 19 Aug 2003 14:35:55 -0400, wrote:

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

On Tue, 19 Aug 2003 23:34:07 -0400, wrote:

Good day Richard,

You have picked an example that simply has different representations
for power. I do not believe there has been any dispute about whether
conversions between different units of power are valid; they are.

The general question is: if two things can be simplified to the same
set of units are they the same thing.
At least two counter examples have been offerred to demonstrate that
just because two things have the same units, they are not the same.

Torque is not work; though they both have N-m as their units.

If you take a solid axle, fix it at one end and twist at the other,
Torque is the plastic deformation in the form of that twist being
distributed along the length of the axle as shearing stress. That
twist allows for some rotation at the end where the rotational force
is applied and that is obviously work. I know, I've calibrated 100's
of Torque wrenches (mostly micrometer click wrenches) from 15 pound-in
to a 600 pound-ft and broke a bench doing it.

Modulus of elasticity is not stress; though they are both expressed
as Pascals (after simplification).

This has a close association with your observation above, so I will
continue with the same model. But first, the definitions that you
seem to accept, but tied into this discussion. From "University
Physics," Sears and Zemansky, containing a chapter called "Elasticity"
whose second section is titled "Stress" (the first section is titled
"Introduction").

"Stress is a force per unit area."

"Strain. ...refers to the relative change in dimensions ... subjected
to stress." As this is distance over distance, strain has no
dimension (the units cancel as has been pointed out by others).

"Elastic modulus. The ratio of a stress to the corresponding strain
is called an elastic modulus. ... Since a strain is a pure number,
the units of Young's modulus are the same as those of stress, namely,
force per unit area. Tabulated values are usually in lb/in² or
dynes/cm²."

Returning to that same axle. We score a line along its length from
free end to fixed end with a scribe that travels a path parallel to
the axis. We apply some force, hold it, and scribe a second line. We
go to the middle of its length and scribe two lines around the
circumference of the axle (a short distance apart). These last two
lines describe opposite shears due to torsion. The stress varies as a
function of depth into the axle (greater at the periphery, less in the
interior). We then examine the enclosed area which describes a twist
per length (area for the applied force - stress). The axial lines are
parallel to the compression and the circumferential lines are parallel
to the tension. If this axle were made of wood, it would fail under
compression when its elasticity was pushed beyond its limit. In
comparison, it would also exhibit a larger rotational displacement
with the same force applied to an iron axle.

The last observation is simply reduced, or normalized as described
above in the definition of modulus and what you obtain for the two
materials is either a constant force with different rotational
displacements (and different scribed areas); or the same rotational
displacement (constant scribed areas) with different applied forces.
You still have torsion, you still have stress and strain, and you
still have rotational displacement - the only difference is in the
material's characteristic which is described by the modulus.


This seems sufficient to prove that two things with the same units
are not necessarily the same.

It proves you have two different materials which is the point of a
modulus in any discipline.


It leaves open the question as to how does one know whether two
things with the same units are the same (or not); a much more
challenging problem, I suspect.

...Keith

Hi Keith,

The only question is what is different, the why follows from fairly
obvious implications of being material based. Your problem is in the
definition of the application of the terms, not their expressions.
You might want to consult a slim volume called
"Elements of Strength of Materials," Timoshenko.

You will note that this bears no relation to ohms being different,
because as you observed with the horsepower example, it is simply
flipping through translations until you hit the units you want.

73's
Richard Clark, KB7QHC

"Dr. Slick" wrote:

wrote in message ...
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

Richard Clark wrote in message . ..


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

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

wrote in message ...
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

On Thu, 21 Aug 2003 06:31:45 -0400, wrote:


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

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

On Tue, 19 Aug 2003 14:47:58 -0700, Roy Lewallen
wrote:

I find this most interesting. As a P.E. licensed by the state of Oregon
(since 1981), I'm aware that I'm subject to state laws governing the
code of conduct of Professional Engineers, and all other applicable
state laws. I didn't realize that I had legal obligations to NIST, or
that any other federal agency has requirements for P.E.s of all states.
Would you please provide some reference where I can further research
this obligation and the rules it has imposed that I'm legally required
to comply with?

Roy Lewallen, W7EL, P.E.

Hi Roy,

"RCW 19.94.150
Standards recognized.
The system of weights and measures in customary use in the United
States and the metric system of weights and measures are jointly
recognized, and either one or both of these systems shall be used
for all commercial purposes in this state. The definitions of
basic units of weight and measure and weights and measures
equivalents, as published by the national institute of standards
and technology or any successor organization, are recognized and
shall govern weighing or measuring instruments or devices used in
commercial activities and other transactions involving weights and
measures within this state."

This is from the state of Washington, I will leave it to you to
research your own particular point of liability in Oregon.

I would add what the IEEE offers into the matter of observing
standards in the development of software and confirming your
disclaimers with:

"The Legal Standard of Professionalism"

"One curious fact from the legal perspective decries a serious
lack: there is no such thing as software malpractice. Why?
A peek into the legal mind provides a disturbing explanation.
There is insufficient evidence to show that programmers
know how to learn from each other, much less from the rest of
the world."

I, for one, could envision you having interest in both, but as I
stated before, I could not see you bothered with the first - seeing
that you have not volunteered any additional details of your trade
aside from software, that stands to good reason.

73's
Richard Clark, KB7QHC

Richard Clark wrote:
On Tue, 19 Aug 2003 14:47:58 -0700, Roy Lewallen
wrote:


I find this most interesting. As a P.E. licensed by the state of Oregon
(since 1981), I'm aware that I'm subject to state laws governing the
code of conduct of Professional Engineers, and all other applicable
state laws. I didn't realize that I had legal obligations to NIST, or
that any other federal agency has requirements for P.E.s of all states.
Would you please provide some reference where I can further research
this obligation and the rules it has imposed that I'm legally required
to comply with?

Roy Lewallen, W7EL, P.E.


Hi Roy,

"RCW 19.94.150
Standards recognized.
The system of weights and measures in customary use in the United
States and the metric system of weights and measures are jointly
recognized, and either one or both of these systems shall be used
for all commercial purposes in this state. The definitions of
basic units of weight and measure and weights and measures
equivalents, as published by the national institute of standards
and technology or any successor organization, are recognized and
shall govern weighing or measuring instruments or devices used in
commercial activities and other transactions involving weights and
measures within this state."

This is from the state of Washington, I will leave it to you to
research your own particular point of liability in Oregon.

Wow, thanks for the heads-up. I'll be more careful to specify circuit
board trace line widths in furlongs, and volumes of radar detection
regions in bushels, those being duly recognized customary units of
measure here in Oregon. I'll no longer use lakj;ofs and mapeurqak!pys,
which I had previously been using.

I would add what the IEEE offers into the matter of observing
standards in the development of software and confirming your
disclaimers with:

"The Legal Standard of Professionalism"

"One curious fact from the legal perspective decries a serious
lack: there is no such thing as software malpractice. Why?
A peek into the legal mind provides a disturbing explanation.
There is insufficient evidence to show that programmers
know how to learn from each other, much less from the rest of
the world."

I, for one, could envision you having interest in both, but as I
stated before, I could not see you bothered with the first - seeing
that you have not volunteered any additional details of your trade
aside from software, that stands to good reason.

As I'm afraid so often happens with your postings, I haven't a clue what
you're trying to say. It sounds vaguely like a complaint, but I can't
for the life of me fathom what about, except that it seems to be some
sort of objection to the legal disclaimers which accompany my software.
Could you please try to rephrase it in a way that can be understood by
an engineer with a sadly deficient liberal arts education?

If you feel that the legal disclaimers which accompany my software are
unduly restrictive or otherwise too onerous for you, or you're not fully
satisfied with EZNEC in any way, all you need do is so state in
peasant-level plain language so I can understand it, and I'll promptly
refund the full purchase price. Just as it says clearly in the EZNEC
manual (Help/Contents/Introduction/Guarantee).

Roy Lewallen, W7EL

On Thu, 21 Aug 2003 20:03:41 -0700, Roy Lewallen
wrote:

Richard Clark wrote:
On Tue, 19 Aug 2003 14:47:58 -0700, Roy Lewallen
wrote:


I find this most interesting. As a P.E. licensed by the state of Oregon
(since 1981), I'm aware that I'm subject to state laws governing the
code of conduct of Professional Engineers, and all other applicable
state laws. I didn't realize that I had legal obligations to NIST, or
that any other federal agency has requirements for P.E.s of all states.
Would you please provide some reference where I can further research
this obligation and the rules it has imposed that I'm legally required
to comply with?

Roy Lewallen, W7EL, P.E.


Hi Roy,

"RCW 19.94.150
Standards recognized.
The system of weights and measures in customary use in the United
States and the metric system of weights and measures are jointly
recognized, and either one or both of these systems shall be used
for all commercial purposes in this state. The definitions of
basic units of weight and measure and weights and measures
equivalents, as published by the national institute of standards
and technology or any successor organization, are recognized and
shall govern weighing or measuring instruments or devices used in
commercial activities and other transactions involving weights and
measures within this state."

This is from the state of Washington, I will leave it to you to
research your own particular point of liability in Oregon.

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.

....
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

Richard Clark wrote:
On Thu, 21 Aug 2003 20:03:41 -0700, Roy Lewallen
wrote:
...

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

On Sun, 17 Aug 2003 17:14:21 GMT, Richard Clark
wrote:

On Sun, 17 Aug 2003 13:56:11 GMT, tad danley
wrote:

Roy Lewallen wrote:


I've tried to point out on this thread that although the feedpoint
impedance is an impedance with the units of ohms, and the impedance of a
plane wave in free space also has the units of ohms, they're not the
same thing.


This may not be a good analogy, but Specific Impulse of rocket motors
helps me to remember that the 'units' of something have to be considered
in the context of what is being measured. Specific impulse is a measure
of the performance of a rocket motor. It measures the thrust obtained
from a single kilogram of propellant burned in one second. The 'units'
of Specific Impulse are seconds, but we're not measuring 'time'.

73,

Hi Tad,

Your point is well taken. ALL physical phenomenon can be expressed
through a chain of conversions in the MKS system of units. When
someone tells you that their terminology is inconsistent between
disciplines (as such offered in this and other threads); it must then
be amenable to reduction to MKS terms or one of the two conflicting
expressions is invalid.

That is to say to the specific matter about the usage of "ohms:"
Here, the unit of ohm must be reduced to Meters, Kilograms, and
Seconds for both usages (electrical and radiative). At that point,
both will have a common basis for comparison and if in fact their
reduced terms are identical, then their common usage is also
identical.

One simple example is with the measurement of body weight on the
bathroom scale (a torsion or compression device) as opposed to the
weight measured on the doctor's scale (a beam balance). Let's say
before you go to the doctor's, you weigh yourself in around 165
pounds. When you arrive at the doctor's, his scale says you weigh
around 75 kilograms.

Let's remove this same scenario to the moon (you live in one of those
futuristic 1990's colonies forecast by the space race back in the
60's). Before you went to the doctor's you weighed in around 33
pounds. When you arrive at the doctor's, his scale says you weigh
around 75 kilograms.

Here we find the expression "pounds" suffers what appears to be the
same plight of "ohms" in that the determination of a value is
inconsistent. You may also note constants of proportionality on earth
and the moon. These constants when expressed as a ratio also describe
the significant differences between the earth and the moon.

The problem is that the term "weight" has a hidden association to the
constant of Gravity. The expression Gram is one of Mass, not weight.
The expression pound is not an expression of Mass unless you expand it
to include the term for the particular constant of Gravity. Mass is
constant in the Newtonian Universe, and weight is not.

If you were to have reduced the pounds to the MKS system both times,
you would have found it consistent both times (here on earth, and on
the moon).

Apparently you are claiming that pounds are not units of mass.

Where did you learn that?

Being the skeptic that I am, how can I convince myself that that is
true? Is there some textbook, or something from some national
standards agency, that would help me verify this?

Gene Nygaard

On Sun, 17 Aug 2003 13:56:11 GMT, tad danley
wrote:

Roy Lewallen wrote:


I've tried to point out on this thread that although the feedpoint
impedance is an impedance with the units of ohms, and the impedance of a
plane wave in free space also has the units of ohms, they're not the
same thing.


This may not be a good analogy, but Specific Impulse of rocket motors
helps me to remember that the 'units' of something have to be considered
in the context of what is being measured. Specific impulse is a measure
of the performance of a rocket motor. It measures the thrust obtained
from a single kilogram of propellant burned in one second. The 'units'
of Specific Impulse are seconds, but we're not measuring 'time'.

It is a bad analogy--for the simple fact that in SI, the proper units
of specific impulse are newton seconds per kilogram (N·s/kg), or the
equivalent meters per second (m/s).

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

On Wed, 24 Sep 2003 16:45:38 GMT, Gene Nygaard
wrote:


Apparently you are claiming that pounds are not units of mass.

Where did you learn that?

Being the skeptic that I am, how can I convince myself that that is
true? Is there some textbook, or something from some national
standards agency, that would help me verify this?

Gene Nygaard

Hi Gene,

Exactly. Perhaps you should re-consider the simple illustration of
difference that I offered in the post you responded to.

Does the weight you measure on a bathroom scale change from the earth
to the moon because your mass changed too? Jenny Craig would have an
armada of shuttles warming up in Florida to a steady trade if that
were true.

However, you do ask for a reference and acknowledge the NIST as a
reputable source (many here ignore this commonplace):
http://physics.nist.gov/PhysRefData/...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

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

Richard Clark wrote:

On Wed, 24 Sep 2003 16:45:38 GMT, Gene Nygaard
wrote:


Apparently you are claiming that pounds are not units of mass.

Where did you learn that?

Being the skeptic that I am, how can I convince myself that that is
true? Is there some textbook, or something from some national
standards agency, that would help me verify this?

Gene Nygaard


Hi Gene,

Exactly. Perhaps you should re-consider the simple illustration of
difference that I offered in the post you responded to.

Does the weight you measure on a bathroom scale change from the earth
to the moon because your mass changed too? Jenny Craig would have an
armada of shuttles warming up in Florida to a steady trade if that
were true.

However, you do ask for a reference and acknowledge the NIST as a
reputable source (many here ignore this commonplace):
http://physics.nist.gov/PhysRefData/...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

On Wed, 24 Sep 2003 17:41:18 GMT, Dave Shrader
wrote:

I weigh approximately 40# on the moon!!! Too skinny for my height 5"9".

I better stay here. But my doctor wants me to loose 40#. Something's
wrong. I need a Twinkie!!

DD

Hi OM,

If you huff down a package of Ex-Lax you would take care of the
doctor's advice with a lot of "loose" weight. (Language is fun ;-)

73's
Richard Clark, KB7QHC

On Wed, 24 Sep 2003 17:23:12 GMT, Richard Clark
wrote:

On Wed, 24 Sep 2003 16:45:38 GMT, Gene Nygaard
wrote:


Apparently you are claiming that pounds are not units of mass.

Where did you learn that?

Being the skeptic that I am, how can I convince myself that that is
true? Is there some textbook, or something from some national
standards agency, that would help me verify this?

Gene Nygaard

Hi Gene,

Exactly. Perhaps you should re-consider the simple illustration of
difference that I offered in the post you responded to.

Does the weight you measure on a bathroom scale change from the earth
to the moon because your mass changed too? Jenny Craig would have an
armada of shuttles warming up in Florida to a steady trade if that
were true.

So what happens when you get serious about your weight and go to the
doctors office or the gym and weigh yourself on one of those platform
type beam balances?

Would your pounds be different on the moon? By how much?

However, you do ask for a reference and acknowledge the NIST as a
reputable source (many here ignore this commonplace):
http://physics.nist.gov/PhysRefData/...constants.html

There is absolutely nothing about pounds on this page. So don't be
bull****ting us.

The link:

is quite specific to the matter.

Not a link directly on the page above; maybe on one of the links
there.

There is absolutely nothing about pounds on this page either. You are
still bull****tiing.

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

On Wed, 24 Sep 2003 18:22:24 GMT, Gene Nygaard
wrote:

On Wed, 24 Sep 2003 17:23:12 GMT, Richard Clark
wrote:

On Wed, 24 Sep 2003 16:45:38 GMT, Gene Nygaard
wrote:

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.

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/...constants.html

There is absolutely nothing about pounds on this page. So don't be
bull****ting 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).

The link:

is quite specific to the matter.

Not a link directly on the page above; maybe on one of the links
there.

There is absolutely nothing about pounds on this page either. You are
still bull****tiing.

Have you tried loosing weight?

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

On Wed, 24 Sep 2003 19:54:06 GMT, Richard Clark
wrote:

On Wed, 24 Sep 2003 18:22:24 GMT, Gene Nygaard
wrote:

On Wed, 24 Sep 2003 17:23:12 GMT, Richard Clark
wrote:

On Wed, 24 Sep 2003 16:45:38 GMT, Gene Nygaard
wrote:

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.

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/...constants.html

There is absolutely nothing about pounds on this page. So don't be
bull****ting 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.

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?

The link:

is quite specific to the matter.

Not a link directly on the page above; maybe on one of the links
there.

There is absolutely nothing about pounds on this page either. You are
still bull****tiing.

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 ;-)

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/

On Wed, 24 Sep 2003 20:15:51 GMT, Gene Nygaard
wrote:


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).

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/...constants.html

There is absolutely nothing about pounds on this page. So don't be
bull****ting 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.

Show me something from NIST saying that pounds are not units of mass.
Or from some textbook.

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.

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 ;-)

I suppose that is an affirmative.

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.

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.

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 wrote:

[SNIP]


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



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