RadioBanter - View Single Post

Stagger Lee

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.