On Tue, 07 Nov 2006 23:08:19 -0500, wrote:
: On Wed, 8 Nov 2006 03:28:13 +0000 (UTC), Stagger Lee
: wrote:
:
: On Tue, 07 Nov 2006 21:48:10 -0500, wrote:
: : On Tue, 7 Nov 2006 23:47:02 +0000 (UTC), Stagger Lee
: : wrote:
: :
: : On 6 Nov 2006 16:31:39 -0800, N9OGL wrote:
: : :
: : : Steve, I suggest you read Part 15, Part 15 has nothing to do with power
: : : (in Wattage) it has to do with the field strength (maininly the
: : : electrical field of the electromaganetic wave) you can put out 100
: : : watts and still have the field strength as prescribed under part 15 (on
: : : 13.556 MHz it 15,848 microvolts per meter at a distance of 30 meters
: : : (90 Feet) CB radio's which run 4 - 5 watts is 10,000 microvolts per
: : : meter at 30 meters.....but you wouldn't know anything about that now
: :
: : Odd. If you take the formula for an isotropic radiator emitting P
: : watts and solve it for the electric field, e, you come up with
: : e = sqr(30*P)/r at a distance of "r" meters (see my earlier posts).
: :
: : what is odd is that you keep leaving the apporoate contants that
: : adjust for the presence of AIR (amoug other thing
:
: Air and vacuum have nearly identical permittivity and permeability -
: the properties which control the impedance of air and of free space
: and which in turn relate electric field strength to the spatial power
: density. It sounds to me as though you are simply blowing smoke and
: have no real understanding of what you just attempted to write.
:
: you are the one without any idea idea of what you are writting
:
: you simply have NO understanding of Gausses law which deteremines
: electircal feild strength

Wrong. In the case of an electromagnetic field, Gauss' law is only
one of four equations (Maxwell's equations) that are needed to
determine the electric field strength. Since a changing magnetic
field can produce an electric field too (Faraday's law), Gauss' law is
not enough. The only time it is sufficient is in the electrostatic
case.

Fortunately, the case under discussion doesn't require the solution of
a set of partial differential equations (Maxwell's equations), because
the solution has been incorporated into the idea of the impedance of
free space. That includes the permittivity of the medium (as used by
Gauss' law) plus the permeability of the medium, and it allows one to
express the power density in space with an Ohm's law type of equation
that involves only the electric field and the impedance.

That simple equation says that 5 watts of radiated power will produce
an electric field strength which is 40 times larger than that claimed
by N9OGL. That's why I asked him if he'd been consulting with Woger
again.

Perhaps he's been consulting with you instead. You just got an "F"
in electromagnetics.


====================
God said, "div D = rho, div B = 0, curl E = -@B/@t, curl H = J + @D/@t,"
and then there was light.

On Wed, 08 Nov 2006 11:50:55 -0500, wrote:
: One Guass governs eltrostatic and electro dynamic cases as well One
: normaly learns the the electrostatic case case first
:
: FOOLs like you don't bother to learn more
:
: but you have failed to account for the fact in any RF mattter power
: dvideds beteween the elictric and Magnetic and then the 2 regenerate
: each other (with some path loss
:
: in short you can't profrom a calution worth doing without more than
: data than you have and more equation

The result of solving Maxwell's equations for the radiated field is
that the electric field strength in volts / meter can be used to
compute the power density in watts per square meter by means of the
simple equation

P = e^2 / (120*Pi)

120*Pi is the impedance of free space, as obtained from Maxwell's
equations.

Now that happens to be a pretty complete equation, and it can indeed
be used as it stands to figure out which power density, P, corresponds
to a particular r.m.s. electric field, e. You can also solve the
equation for "e" if you want to calculate the electric field when you
already know the power density. It is a steady-state solution, and it
excludes non-radiating fields (the so-called near field), but it is
good enough for most initial calculations.

So, no, one doesn't need additional equations, and one doesn't need
to worry about power division between the electric and magnetic
fields. That's already been included in the formula given above.

: none of which matter either

They only don't matter because you have the details wrong. If we
tried to get at the radiated field strength your way, we'd never get
an answer at all.

: what matters is the measurable filed strangth

If you can't estimate the field strength beforehand, then you CAN'T
possibly come up with a good initial station design. You'd have to
put something up and hope it works out. I suspect that's exactly what
Todd has done, and it is a very dangerous way to operate.

: to determine if Todd is legal or not you MUST prefrom the test and
: till that is done Innocent till found guility is ther Law

If the calculations suggest that Todd is very probably exceeding the
legal electric field strenth (and they do), it is something that no
sane station operator should ignore. Calculations are inexpensive: A
certified field strength meter is not, and a NAL is even costlier.

Saggytits Lee plagiarized and added a few of his own comments to make
it look like he wrote the whole thing again:
On Wed, 08 Nov 2006 11:50:55 -0500, wrote:
: One Guass governs eltrostatic and electro dynamic cases as well One
: normaly learns the the electrostatic case case first
:
: FOOLs like you don't bother to learn more
:
: but you have failed to account for the fact in any RF mattter power
: dvideds beteween the elictric and Magnetic and then the 2 regenerate
: each other (with some path loss
:
: in short you can't profrom a calution worth doing without more than
: data than you have and more equation

The result of solving

Saggytits plagiarizes some more. Poor Saggytits.

On Wed, 08 Nov 2006 21:32:30 -0500, wrote:
[nothing useful]

I see that you need a great deal of expert instruction. Fortunately,
I happen to have some time available this month, so I will attempt to
enlighten you on the subject of Maxwell's equations and electromagnetic
radiation.

In my absence, please feel free to consult either Todd Daugherty or
Roger Wiseman.

First assignment: Define a field as used in physics, and then try to
describe both static magnetic and static electric fields. Feel free
to use Wikipedia if you get stuck. Later on, we'll move to dynamic
fields, because they form the basis of electromagnetic radiation.

P.S. Maxwell's equations have appeared in my sig several times (see
the current sig), so you were quite wrong when you said that I
couldn't write them down properly. What happened instead is that you
simply don't know enough to recognize Maxwell's equations when they
are right in front of your face. If you can't recognize Maxwell's
equations, how in the world will you ever pass the new, code-free
Extra class exam? Think about it.

====================
God said, "div D = rho, div B = 0, curl E = -@B/@t, curl H = J + @D/@t,"
and then there was light.

Stagger Lee wrote:
[nothing useful since he plagiarized the bulk of his post from another website]

"****UP Lee" geography idiot and imbecile
****s up as usual when he thinks Marshall County, Alabama
and Marshall County, West Virginia are the same place in message
:
" Then what is at http://www.geocities.com/skywarncanwarn/ doodle
brain? It sure looks like the Marshall County ARES, and it even has a
place to
send email."

"Not Cocksucker Lloyd" wrote in message
oups.com...

Stagger Lee wrote:
[nothing useful since he plagiarized the bulk of his post from another
website]

Prove it, Woger. Since you are the Prince of Plagiarists you automatically
assume others do as you.

As you used to say, ppostfu.