Richard Clark wrote:
kg/(s·m²)

One needs to multiply by the area under consideration
to obtain the total momentum which is conserved.

It's the same as multiplying the Poynting vector
by the area under consideration to obtain the
total energy.
--
73, Cecil http://www.w5dxp.com

On Thu, 22 Nov 2007 05:20:10 -0600, Cecil Moore
were Cecil's dignity was not conserved, wrote:
If we carry out the math of your own answer, it renders units in your
terms to:
(watts/m²)/(m/s)²
or
((kg·m²/s³)/m²)/(m/s)²
or
kg/(s·m²)

One needs to multiply by the area under consideration
to obtain the total momentum which is conserved.

Which, of course, yields:
kg/s
which is not momentum, but oddly enough the units one might use in
describing how fast his bath tub was filling with photons. :-)

Oh well, third time's a charm! Keep fumbling with the conservation of
dignity, and eventually you might bumble into a job with Anderson
Consulting doing Enron's books.

On Thu, 22 Nov 2007 07:34:14 -0800, Richard Clark
wrote:

On Thu, 22 Nov 2007 05:20:10 -0600, Cecil Moore
were Cecil's dignity was not conserved, wrote:
If we carry out the math of your own answer, it renders units in your
terms to:
(watts/m²)/(m/s)²
or
((kg·m²/s³)/m²)/(m/s)²
or
kg/(s·m²)

One needs to multiply by the area under consideration
to obtain the total momentum which is conserved.

Which, of course, yields:
kg/s
which is not momentum, but oddly enough the units one might use in
describing how fast his bath tub was filling with photons. :-)

Oh well, third time's a charm! Keep fumbling with the conservation of
dignity, and eventually you might bumble into a job with Anderson
Consulting doing Enron's books.

Hi All,

Well, no point in waiting for another rationalization when I am
perfectly capable of filling that in for Cec' (and more entertaining
than him when I do).

Let's see if I can pull together a good old-boy drawl and a scrub of
the boot toe in the dirt:
"One needs to multiply by the volume under consideration
to obtain the total momentum which is conserved!"

Umm, yes, if your Xeroxed authors need that much help in you
describing what they must have meant, but didn't say, then throwing in
previously undisclosed terms might do the trick.

However, looking aside from this obvious self-serving manipulation of
the books (Anderson would be proud, in a perverted sense) it then
gives us a third dimension of meters (the one you will have suckered
into the equation) which is also a time specification (and this would
then be called not Momentum, but Impulse, which does carry the same
units but is the Integration over time).

"Yes! Of course! This is called the Conservation of Impulse!
It is EXACTLY what my references meant to say...."

And then we return to an observation I made earlier about:
The short journey was described by the term "dt".

Ah, suffering the dt's.

As Ed McMahon would prompt Johnny:
"Just how short was that journey?"

My guess it will either be too short to do the job, or much too large
to be true.
and we had been left holding the bag once again with asking how big dt
is? Which, of course, also flummoxed Cecil (his having not yet had
the epiphany of what his references "meant to say" but left unsaid).
Some might begin to wonder how they earned a salary in the career of
teaching.

And to whip a dead horse, I also said:
This thread should be called:
"Supposition"
or
"Imposition"
or
"Superstition"

To the group,

Sorry for having unleashed yet another law of conservation that will
undoubtedly yield 100s of postings of "proofs" that in and of
themselves will actually contain no intellectual nourishment.

Next week:
"The Conservation of Radiation Buoyancy"

73's
Richard Clark, KB7QHC

Richard Clark wrote:
Umm, yes, if your Xeroxed authors need that much help in you
describing what they must have meant, but didn't say, then throwing in
previously undisclosed terms might do the trick.

I probably misspelled a word also, Richard, so you can
also jump on that with all four feet. Of course, it should
have been "volume" instead of "area". It's a mental mistake
that is easy to make and it certainly not the same magnitude
of your mistake of declaring that reflections from non-
reflective glass are brighter than the surface of the sun.
--
73, Cecil http://www.w5dxp.com

Richard Clark wrote:
Which, of course, yields:
kg/s
which is not momentum, ...

You made a mistake somewhere, Richard. The equation I
gave is a *volume density, not an area density* so you
are one 'm' short. You should have gotten kg*m/s
--
73, Cecil http://www.w5dxp.com