"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