"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

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

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 ...

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

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



But it's still dimensionless, a "pure number" as they call it.



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


In context, i would agree, but they are still just pure numbers.





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

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


ok, length of angle divided by length of radius, that's right.


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


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


Actually, thats 2*pi*radius*force*moment arm. Right.



...Keith

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

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



Your point it well taken, but cancellation of units has always
worked for me, and everyone else i went to college and high school
with.

And I admit that a wave traveling in a transmission line is
different from an EM wave traveling through space. All i'm saying is
that the E field is defined by a voltage potential field, and the H
field by amperes, so to say that the E field has nothing to do with
voltage potential is a wrong statement in my opinion. And it's still
ohms for the impedance.




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

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

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

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

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

...Keith


It's a difficult question, and i'm glad we are discussing it. All
i'm saying is, the units have to be the same, where ever you use them.


Slick

"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