for the tin hat folks
JC Dill
--
"The nice thing about a mare is you get to ride a lot
of different horses without having to own that many."
~ Eileen Morgan of The Mare's Nest, PA
Jeff Liebermann
>http://www.lessemf.com/personal.html
Chuckle. I thought silver lame went out with Liberace and Elvis.
What I find amusing is that they seem to think that metal woven fabric
is proper protection against RF exposure. It doesn't work that way.
What the metal does is redirect the RF. One would assume that it all
gets reflected but that's not the case. What happens is that unless
the garment totally encloses the wearer, any gaps or seams will have a
potential difference across them. Instead shielding, a garment with a
gap acts like a receiving antenna. The large gap across the open
front of the shirt will have a rather substantial field across it,
much of which will be re-radiated into the wearer. Similarly, if the
head is unprotected, the field across the collar would be much larger
than by a direct exposure.
To be truly effective, the cloth should absorb, not reflect, RF
radiation. This can easily be done with carbon doped fibers and other
military anti-reflective technology. For the ultimate in shielding,
ferrite materials can be used to magnetically redirect the signal
along an absorbent surface by Faraday rotation. I guess looking like
a stealth bomber is not currently fashionable.
Well, maybe they're not all that clueless. See the "microwave
absorbing sheet" at:
http://www.lessemf.com/fabric.html
--
Jeff Liebermann je...@comix.santa-cruz.ca.us
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558
Tim May
Liebermann <je...@comix.santa-cruz.ca.us> wrote:
> JC Dill <jcd...@gmail.com> hath wroth:
>
> >http://www.lessemf.com/personal.html
>
> Chuckle. I thought silver lame went out with Liberace and Elvis.
>
> What I find amusing is that they seem to think that metal woven fabric
> is proper protection against RF exposure. It doesn't work that way.
> What the metal does is redirect the RF. One would assume that it all
> gets reflected but that's not the case. What happens is that unless
> the garment totally encloses the wearer, any gaps or seams will have a
> potential difference across them. Instead shielding, a garment with a
> gap acts like a receiving antenna. The large gap across the open
> front of the shirt will have a rather substantial field across it,
> much of which will be re-radiated into the wearer. Similarly, if the
> head is unprotected, the field across the collar would be much larger
> than by a direct exposure.
This is wholly incorrect.
Wave physics does not work this way. If 10 m wavelength water waves hit
a jetty, a 1 m gap does NOT seen any intensification of the wave
strength. (Far from it, it is effectively fully shielded from waves
which are a few times or more longer in wavelength than itself, it's
gap.)
If you argue, perhaps, that EM radiation is somehow different from
water waves, you are even more ignorant than I assumed.
(The differences are significant when the wave vs. photon issue
becomes significant, usually only measurable at visible wavelengths and
shorter, and never observed at RF wavelengths.)
Having worked inside Faraday cages, I know first-hand, confirming
obvious EM theory, that a small gap does not "intensify" entering
radiation.
(The key is that conductors cannot support electric fields. This makes
most conductor "reflective." The radiation not reflected, usually away
from the emitter, can be aborbed by a series of internal baffles and
ferrite materials. But for the F-117A and B-2, the biggest win is in
reflecting away incoming radar signals.
(Easily defeated by suitably coordinated ground-based radar, a point
made by some friends of mine in 1988, showing how "Stealth" could be
defeated. They faced pressure from DoD to shut the fuck up. Welcome to
America.)
> To be truly effective, the cloth should absorb, not reflect, RF
> radiation. This can easily be done with carbon doped fibers and other
> military anti-reflective technology. For the ultimate in shielding,
> ferrite materials can be used to magnetically redirect the signal
> along an absorbent surface by Faraday rotation. I guess looking like
> a stealth bomber is not currently fashionable.
The F-117A works largely with flat surfaces, reflecting radar signals
away. Reflecting is more useful than absorbing. A scattered signal is
as good as an absorbed signal, and a lot easier to achieve. This
reduces the "return" signal (sort of the opposite of a corner
reflector).
Everytime you post about these subjects, I see a former "A/V nerd"" who
learned something about how to solder and run wires but who missed out
out on freshman physics, let alone 3rd year electromagnetic theory, let
alone Jackson's "Classical Electrodynamics" and the like.
"Jeff, you're needed in the 9th grade Earth Science classroom for a
showing of "Our Natural Waterfalls.""
--Tim May
Jeff Liebermann
>In article <ndg332t7q29f9gl8t...@4ax.com>, Jeff
>Liebermann <je...@comix.santa-cruz.ca.us> wrote:
>
>> JC Dill <jcd...@gmail.com> hath wroth:
>>
>> >http://www.lessemf.com/personal.html
>>
>> Chuckle. I thought silver lame went out with Liberace and Elvis.
>>
>> What I find amusing is that they seem to think that metal woven fabric
>> is proper protection against RF exposure. It doesn't work that way.
>> What the metal does is redirect the RF. One would assume that it all
>> gets reflected but that's not the case. What happens is that unless
>> the garment totally encloses the wearer, any gaps or seams will have a
>> potential difference across them. Instead shielding, a garment with a
>> gap acts like a receiving antenna. The large gap across the open
>> front of the shirt will have a rather substantial field across it,
>> much of which will be re-radiated into the wearer. Similarly, if the
>> head is unprotected, the field across the collar would be much larger
>> than by a direct exposure.
>
>This is wholly incorrect.
>
>Wave physics does not work this way. If 10 m wavelength water waves hit
>a jetty, a 1 m gap does NOT seen any intensification of the wave
>strength. (Far from it, it is effectively fully shielded from waves
>which are a few times or more longer in wavelength than itself, it's
>gap.)
>
>If you argue, perhaps, that EM radiation is somehow different from
>water waves, you are even more ignorant than I assumed.
Sigh. About 8 years ago, I was trying to figure out why one of the
tower tech's at commercial site was claiming that he was getting
zapped by his aluminized RF protective suit, but only on some active
towers. I had originally assumed that it was some type of resonant
effect where the suit formed a cavity or something. Nope. The chest
diameter of the suit was about 1 meter around, which forms a rather
nice half wave loop antenna if the front seams are left open. With a
nearby 300 watt VHF paging xmitter, there was quite a bit of voltage
across the gap. It seemed this tower climber was not closing the
front of the suit (it's hot in there) and causing the problem. When
the metallic Velcro "zipper" was closed, no problem. Since the wearer
was feeling the sparks, I assume that the field was being re-radiated
from the gap.
It's something like a Hertzian receive loop used for spark gap
reception:
| http://www.fas.harvard.edu/~scidemos/ElectricityMagnetism/HertzResonator/HertzResonator.html
where the chest cavity forms the loop, and the open front forms the
gap.
The RF protective suit vaguely resembles the protective shirts offered
by the RF paranoia web site. Just one problem. The front seams are
not closed. Depending on the frequency, it may also have the
aforementioned gap problem.
>(The differences are significant when the wave vs. photon issue
>becomes significant, usually only measurable at visible wavelengths and
>shorter, and never observed at RF wavelengths.)
The EMF protective shirts are advertised as protective from 10KHz to
1GHz. 1Ghz is a wavelength of about 30cm which will easily penetrate
any gap larger than about 7.5cm.
>Having worked inside Faraday cages, I know first-hand, confirming
>obvious EM theory, that a small gap does not "intensify" entering
>radiation.
I didn't say "intensify". I said:
"The large gap across the open front of the shirt will
have a rather substantial field across it, much of which
will be re-radiated into the wearer."
There will be losses involved. There's no such thing as a static RF
field. The conducted RF has to go somewhere. Much of it will be
reflected, some of it will be dissipated in losses across the gap, and
some of it will be re-radiated to the inside of the shield.
>(The key is that conductors cannot support electric fields. This makes
>most conductor "reflective." The radiation not reflected, usually away
>from the emitter, can be aborbed by a series of internal baffles and
>ferrite materials. But for the F-117A and B-2, the biggest win is in
>reflecting away incoming radar signals.
I note that the surface of these stealth airplanes is NOT silver lame
or other mirror like finish. It's a powdered iron and ferrite based
paint designed to induce losses at RF frequencies.
http://www.aeronautics.ru/f117a.htm
"...the coating contains carbonyl iron ferrite (special paint
using this material is known as "iron ball" paint). When a radar
wave encounters this coating, it creates a magnetic field within
the metallic elements of the coating. The field has alternating
polarity and dissipates the energy of a radar signal. A significant
portion of radar energy is converted into heat."
This is the way methinks a proper RF protective shirt *SHOULD* work.
>(Easily defeated by suitably coordinated ground-based radar, a point
>made by some friends of mine in 1988, showing how "Stealth" could be
>defeated. They faced pressure from DoD to shut the fuck up. Welcome to
>America.)
Indirect illumination radar has been around since the late 1960's. CW
(continuous wave) illlumination with interferometry will do an
adequate job of detection. There's even a system that works using
broadcast FM and TV transmitters, where any disturbance in the field
can be detected and located.
The article also mentions:
http://www.aeronautics.ru/f117a.htm
"This means that to effectively track F-117A one would have to use
multiple radars (or, at least, multiple receivers.)"
Incidentally, stealth aircraft have been detected by the radar
signature of the rather massive air turbulence created by their
non-airodynamic design using simple water absorption doppler radar.
The idea is to bounce the signal off the water vapor, and not the
aircraft.
>The F-117A works largely with flat surfaces, reflecting radar signals
>away. Reflecting is more useful than absorbing. A scattered signal is
>as good as an absorbed signal, and a lot easier to achieve. This
>reduces the "return" signal (sort of the opposite of a corner
>reflector).
If the RF protective shirt was purely reflective (silver lame), then
much of the reflection would end up exposing the head, arms, legs, and
such. Reflections from the shoulders will illuminate the head.
Reflections from the chest will illuminate the arms. Although I'll
admit that it might be better than no protection, the lack to total
coverage makes the RF protective shirt far from optimimum. My
guess(tm) is that a reflective shirt, with an unprotected head, will
roughly double the exposure to the head area (sum of incident plus
shoulder reflected power). However, were the shirt absorptive rather
than reflective, there would be no reflective exposure of the head or
extremeties.
Tim May
Liebermann <je...@comix.santa-cruz.ca.us> wrote:
> There will be losses involved. There's no such thing as a static RF
> field. The conducted RF has to go somewhere. Much of it will be
> reflected, some of it will be dissipated in losses across the gap, and
> some of it will be re-radiated to the inside of the shield.
Nope. Consult an undergraduate E&M textbook, such as Corson and
Lorraine, for how a conductor "reflects" EM waves. It does so by a
back-EMF induced in the conductor.
The "RF has to go somewhere" is a complete misunderstanding of what
happens when an EM wave impinges on a conductor.
As for "re-radiating" to the _inside_ of a shield, you are once again
repeating your errors about fields inside conductors. You really need
to spend an hour or so brushing up on Gauss's Law, at the simplest
level, and the divergence (div) role in fields. Coulomb, for example.
(As with a hollow sphere or any other shape, such as an asteroid with a
hollow including the center of mass, there will be no felt
gravitational field in the hollow. This applies to objects as big as
the earth: one would be weightless through a hollow sphere centered on
the center of mass. Because the only "field" one feels is related to
the mass _inside the inner sphere_. All contributions from the various
bits of the earth above, to the side, on the other side of the hollow,
etc., all cancel out exactly (a useful integral calculus exercise to do
at least once, just to see how the inverse square law neatly cancels
out the contributions).
In the case of a conductor, the same is true of EM fields. Inside a
hollow, no fields.
There are two equivalent-outcome ways to look at this:
1. Charges on the conductor move around (redistribute themselves) so as
to cancel out all fields inside the object, including inside hollows,
exactly.
(How fast this happens, how complete this process is, depends on a few
things like the conductivity, dimensions, the time involved, etc. No
one is saying that a relatively crummy electrical conductor, like cast
iron, for example, is going to respond to a 10 GHz EM wave
instantaneously. Skin depth is an important issue. This is why some
conductors of a given thickness and geometry make better EM shields
than other materias, why mu metal handles the magnetic components
better, etc. )
>
> >(The key is that conductors cannot support electric fields. This makes
> >most conductor "reflective." The radiation not reflected, usually away
> >from the emitter, can be aborbed by a series of internal baffles and
> >ferrite materials. But for the F-117A and B-2, the biggest win is in
> >reflecting away incoming radar signals.
>
> I note that the surface of these stealth airplanes is NOT silver lame
> or other mirror like finish. It's a powdered iron and ferrite based
> paint designed to induce losses at RF frequencies.
Silver or "mirror-like" are OPTICAL WAVELENGTH issues.
A completely black-painted piece of metal may be optically the opposite
of a mirror but nevertheless an excellent reflector of radio-frequency
waves.
I cannot believe you think that the fact that Stealth planes are
dark-colored and not mirror-finished has anything whatsoever to do with
RF reflectance.
Jeez, it is really time to get back to some basics.
> http://www.aeronautics.ru/f117a.htm
> "...the coating contains carbonyl iron ferrite (special paint
> using this material is known as "iron ball" paint). When a radar
> wave encounters this coating, it creates a magnetic field within
> the metallic elements of the coating. The field has alternating
> polarity and dissipates the energy of a radar signal. A significant
> portion of radar energy is converted into heat."
>
> This is the way methinks a proper RF protective shirt *SHOULD* work.
You conveniently and disingenuously left out this bit, which confirms
what I said about the faceted shape of the 117A:
"External LO geometry
In 1966 a well-known Soviet mathematician, Pyotr Ufimtsev,
published a paper in which he described mathematical methods to predict
RCS of two-dimensional objects. Ufimtsev's work was directly used by
Lockheed's mathematician, Denys Overholser, to develop computer
software known as "Echo 1", which could calculate the RCS of an
aircraft constructed of flat panels. This program was used to find the
optimum geometry to minimize an aircraft's RCS. The resulting structure
became known as "Hopeless Diamond", which lies in the basis of F-117A's
external construction. Simply put, the flat, angled external panels of
F-117A are designed to reflect radar waves in all directions but the
direction of the radar's receiving antenna."
> >The F-117A works largely with flat surfaces, reflecting radar signals
> >away. Reflecting is more useful than absorbing. A scattered signal is
> >as good as an absorbed signal, and a lot easier to achieve. This
> >reduces the "return" signal (sort of the opposite of a corner
> >reflector).
>
> If the RF protective shirt was purely reflective (silver lame)
Again, you confuse RF shielding (through reflection and absorbtion)
with "silver lame," an OPTICAL WAVELENGTHS matter.
Jeez, to use the other pronunciation of "lame," you really are.
> , then
> much of the reflection would end up exposing the head, arms, legs, and
> such. Reflections from the shoulders will illuminate the head.
> Reflections from the chest will illuminate the arms. Although I'll
> admit that it might be better than no protection, the lack to total
> coverage makes the RF protective shirt far from optimimum. My
> guess(tm) is that a reflective shirt, with an unprotected head, will
> roughly double the exposure to the head area (sum of incident plus
> shoulder reflected power). However, were the shirt absorptive rather
> than reflective, there would be no reflective exposure of the head or
> extremeties.
You grossly overestimate the contributions from reflection to the head.
First, imagine a totally reflective (in optical wavelengths, for this
Gedankenexperiment) body, along the lines of the liquid metal "T2" in
the movie. Ask yourself this: how much of the T2's head can you see
reflected from any part of his body?
From the front view, a very small bit of the chest area will reflect
part of his lower face. From the back view, an even smaller part of the
back of his head, from small parts of the upper spine and small patches
on the shoulders, angled just properly.
From the sides, a bit of the shoulder will reflect a bit of his head.
(A patch, though, comparable in size to the size of the patch, because
of the distances involved.)
In fact, looking at such a liquid metal person, nearly all of the
"scenery" would be of a distored, reflected panorama of the viewers
half of the full 4-pi solid angle. Do the Gedankenexperiment to see
this.
Now, what one is "seeing" is isomorphic to what a radar or other RF
source is "illuminating" (just swap sources for receivers).
In other words, the reflected component to a person's unshielded ("non
reflective") head is a very small fraction of the component going
directly to the head.
(This can be solved exactly with ray-tracing, of course.)
Second, if the shielding is NOT VERY REFLECTIVE, in the specular
reflection sense, but is instead a _scattering_ medium, the same
calculation applies. Think about it.
And there's an even more interesting and general way to look at this:
Any reflecting or scattering surface--such as the liquid metal torso,
legs, and arms of the T2 character--has many, many "places" to "send"
the reflections, no matter the light source.
That is, all of the various people standing around, or cameras looking
at the character, and so on, will see this liquid metal guy. Just as we
did, quite realistically rendered, in "T2."
This says right off the bat that _most_ of the reflected or scattered
illumination (from an optical source in this thought experiment, or
from radar or RF in your "RF shielding clothing" example) IS IN FACT
REFLECTED OR SCATTERED AWAY.
It takes special planning and arrangement of reflectors to get a
significant fraction reflected onto the face or head. (Those special
fold-out reflectors the cancer enthusiasts use at the beach, for
example.)
This "all other things being equal" or "parsimony" approach is a
powerful way of looking at a lot of situations.
But many of your misapprehensions keep coming from a lack of grounding,
so to speak, in the basic physics.
--Tim May
Jeff Liebermann
>Nope. Consult an undergraduate E&M textbook, such as Corson and
>Lorraine, for how a conductor "reflects" EM waves. It does so by a
>back-EMF induced in the conductor.
>
>The "RF has to go somewhere" is a complete misunderstanding of what
>happens when an EM wave impinges on a conductor.
If the shield were contiguous, I would have no problem with your
explanation. However, I'm talking about a shirt with an open front.
Instead of a shield, it's more like a "slot" antenna. Such antennas
were commonly used in the early rocket program for telemetry as they
could be worked nicely into allegedly contiguous metal surface of the
rocket and did not involve any projecting wires. Here's a slot
antenna made from waveguide:
http://ftp.ij.net/packrats/Slot_Antenna/slot.html
Some basics on how it works:
http://www.tpub.com/content/neets/14183/css/14183_163.htm
Note that the metal surface need not be flat and can wrap around as in
the waveguide slot antenna.
My point is around the slot, there is a substantial field. It is not
uniform and therefore does not complete cancel on the inside of the
shield as you so eloquently note. Instead, you have a RF potential
across the gap as in the slot antenna.
I think we can agree that there's no such thing as a "static" RF
field. So what happens to the energy (RF) that appears across the
gap. Well, it gets re-radiated. Again, think of the gap as part of
an antenna, not a shield. If I place a matching loss (i.e. load)
across the gap, it would absorb all the RF energy and nothing would be
re-radiated. However, in this case, there is no matching load. The
termination impedance of this antenna is roughly infinite. In
wireless terms, that creates a reflective termination, which causes
any arriving energy to be reflected back to it's source. In wireless
terminology, this is called a high VSWR mismatch. The exact manner in
which the return signal is re-reradiated or absorbed is dependent on
antenna resonant effects and internal (body) losses.
>As for "re-radiating" to the _inside_ of a shield, you are once again
>repeating your errors about fields inside conductors. You really need
>to spend an hour or so brushing up on Gauss's Law, at the simplest
>level, and the divergence (div) role in fields. Coulomb, for example.
I'm not suggesting that re-radiation couples all the RF uniformly
inside the shirt. I am suggesting that it presents a rather
substantial local hot spot near the gap in front (and near the head
and arm holes).
>(As with a hollow sphere or any other shape, such as an asteroid with a
>hollow including the center of mass, there will be no felt
>gravitational field in the hollow. This applies to objects as big as
>the earth: one would be weightless through a hollow sphere centered on
>the center of mass. Because the only "field" one feels is related to
>the mass _inside the inner sphere_. All contributions from the various
>bits of the earth above, to the side, on the other side of the hollow,
>etc., all cancel out exactly (a useful integral calculus exercise to do
>at least once, just to see how the inverse square law neatly cancels
>out the contributions).
I don't know of any way to conveniently create a discontinuity in a
gravitational field or use gravity to model the RF shielding shirt.
>In the case of a conductor, the same is true of EM fields. Inside a
>hollow, no fields.
>
>There are two equivalent-outcome ways to look at this:
>
>1. Charges on the conductor move around (redistribute themselves) so as
>to cancel out all fields inside the object, including inside hollows,
>exactly.
In an RF field, this distribution of charge is time variant and
subject to resonances. As in coaxial cable, the vector sum of all the
fields, at any time (or phase) is zero. However, cut a slot in the
coaxial shield, and we have a commercial antenna product called
Radiax.
http://www.andrew.com/products/trans_line/radiax/
which is used for "leaky coax" distributed antenna systems.
>(How fast this happens, how complete this process is, depends on a few
>things like the conductivity, dimensions, the time involved, etc. No
>one is saying that a relatively crummy electrical conductor, like cast
>iron, for example, is going to respond to a 10 GHz EM wave
>instantaneously. Skin depth is an important issue. This is why some
>conductors of a given thickness and geometry make better EM shields
>than other materias, why mu metal handles the magnetic components
>better, etc. )
Sure. That's why I suggested that the proper RF radiation exposure
shirt should be made from absorptive material, not reflective. The
tendency of reflective and conductive material to become antennas is
not as prevalent in absorptive material. They can still become (slot)
antennas, but the losses are so high that little RF appears across the
gap.
>> I note that the surface of these stealth airplanes is NOT silver lame
>> or other mirror like finish. It's a powdered iron and ferrite based
>> paint designed to induce losses at RF frequencies.
>
>Silver or "mirror-like" are OPTICAL WAVELENGTH issues.
Yeah, but they're "cool" looking. Just think of the increased public
relations value and corresponding increase in federal funding for the
program if it could be made to look "cool". If the stealth airplanes
were totally dependent on anti-reflective geometry, then the surface
could be made shiny. However, it's not and much of the low radar
signature comes from absorptive effects.
>A completely black-painted piece of metal may be optically the opposite
>of a mirror but nevertheless an excellent reflector of radio-frequency
>waves.
>
>I cannot believe you think that the fact that Stealth planes are
>dark-colored and not mirror-finished has anything whatsoever to do with
>RF reflectance.
My comments were in response to you allegation that:
"The F-117A works largely with flat surfaces, reflecting radar
signals away. Reflecting is more useful than absorbing.
A scattered signal is as good as an absorbed signal, and a
lot easier to achieve. This reduces the "return" signal (sort
of the opposite of a corner reflector)."
By my warped logic, if anti-reflection were as important as what you
claim, then the designers should have foregone any attempts at
absorptive coatings and relied totally on reflection.
Incidentally, it is my marginally informed opinion that it is easier
to implement an absorptive coating, than make an anti-reflective
airplane fly. The F-117a is difficult to fly and is only stable under
continuous computer control. Absorptive coatings have been around
since the early 1960's(?) when the British demonstrated the technology
by overflying the US and never showing up on the US air defense radar
system. The problem is that it's difficult to produce a wide
frequency range anti-reflective coatings.
>Jeez, it is really time to get back to some basics.
Agreed. Please climb out of your Faraday shield room and read up on
the practical applications of some of the theory you mentioned.
>> http://www.aeronautics.ru/f117a.htm
>> "...the coating contains carbonyl iron ferrite (special paint
>> using this material is known as "iron ball" paint). When a radar
>> wave encounters this coating, it creates a magnetic field within
>> the metallic elements of the coating. The field has alternating
>> polarity and dissipates the energy of a radar signal. A significant
>> portion of radar energy is converted into heat."
>>
>> This is the way methinks a proper RF protective shirt *SHOULD* work.
>You conveniently and disingenuously left out this bit, which confirms
>what I said about the faceted shape of the 117A:
Well, yes I did intentionally. I don't like quoting large blobs of
text. I was only interested in mentioning that how the absorptive
coating works, and perhaps speculating on whether it could be used to
RF proof clothing. Probably not, because it would be rather heavy.
>"External LO geometry
> In 1966 a well-known Soviet mathematician, Pyotr Ufimtsev,
>published a paper in which he described mathematical methods to predict
>RCS of two-dimensional objects. Ufimtsev's work was directly used by
>Lockheed's mathematician, Denys Overholser, to develop computer
>software known as "Echo 1", which could calculate the RCS of an
>aircraft constructed of flat panels. This program was used to find the
>optimum geometry to minimize an aircraft's RCS. The resulting structure
>became known as "Hopeless Diamond", which lies in the basis of F-117A's
>external construction. Simply put, the flat, angled external panels of
>F-117A are designed to reflect radar waves in all directions but the
>direction of the radar's receiving antenna."
Well, yes. That explains the origins of the geometry used to design
the stealth aircraft. I was only talking about the absorptive surface
coating.>
>> If the RF protective shirt was purely reflective (silver lame)
>
>Again, you confuse RF shielding (through reflection and absorbtion)
>with "silver lame," an OPTICAL WAVELENGTHS matter.
Ok, I'll try to be serious.
>Jeez, to use the other pronunciation of "lame," you really are.
>You grossly overestimate the contributions from reflection to the head.
http://people.csail.mit.edu/rahimi/helmet/
>First, imagine a totally reflective (in optical wavelengths, for this
>Gedankenexperiment) body, along the lines of the liquid metal "T2" in
>the movie. Ask yourself this: how much of the T2's head can you see
>reflected from any part of his body?
A bit more than can be viewed directly from an equal altitude
position. The reflective component isn't huge, but depending on the
arrival angle of the RF signal, it can be substantial. If the signal
came from above, as from a cell tower, it would be considerable. There
are also reflections from other parts of the shirt shield, that could
eventually end up in the unprotected head. Even if I the reflective
contribution was fairly small, it still would be more than what the
unprotected head would receive from direct exposure. Remember, the
idea of the RF reflective shirt is to protect the user from RF
exposure, not increase it.
Time for some ray tracing?
>From the front view, a very small bit of the chest area will reflect
>part of his lower face. From the back view, an even smaller part of the
>back of his head, from small parts of the upper spine and small patches
>on the shoulders, angled just properly.
>
>From the sides, a bit of the shoulder will reflect a bit of his head.
>(A patch, though, comparable in size to the size of the patch, because
>of the distances involved.)
>
>In fact, looking at such a liquid metal person, nearly all of the
>"scenery" would be of a distored, reflected panorama of the viewers
>half of the full 4-pi solid angle. Do the Gedankenexperiment to see
>this.
The "thank you experiment"? Huh?
>Now, what one is "seeing" is isomorphic to what a radar or other RF
>source is "illuminating" (just swap sources for receivers).
>
>In other words, the reflected component to a person's unshielded ("non
>reflective") head is a very small fraction of the component going
>directly to the head.
>
>(This can be solved exactly with ray-tracing, of course.)
>
>Second, if the shielding is NOT VERY REFLECTIVE, in the specular
>reflection sense, but is instead a _scattering_ medium, the same
>calculation applies. Think about it.
The scattering from an uneven surface would simply even out the hot
spots and the dead spots. The overall bulk exposure would be about
the same.
>And there's an even more interesting and general way to look at this:
>
>Any reflecting or scattering surface--such as the liquid metal torso,
>legs, and arms of the T2 character--has many, many "places" to "send"
>the reflections, no matter the light source.
>
>That is, all of the various people standing around, or cameras looking
>at the character, and so on, will see this liquid metal guy. Just as we
>did, quite realistically rendered, in "T2."
>
>This says right off the bat that _most_ of the reflected or scattered
>illumination (from an optical source in this thought experiment, or
>from radar or RF in your "RF shielding clothing" example) IS IN FACT
>REFLECTED OR SCATTERED AWAY.
T2 didn't have an open slot down the front.
>It takes special planning and arrangement of reflectors to get a
>significant fraction reflected onto the face or head. (Those special
>fold-out reflectors the cancer enthusiasts use at the beach, for
>example.)
>
>This "all other things being equal" or "parsimony" approach is a
>powerful way of looking at a lot of situations.
>
>But many of your misapprehensions keep coming from a lack of grounding,
>so to speak, in the basic physics.
Similarly, you seem to be lacking in the practical application of some
of that basic physics. You're also having trouble staying on the
current subject, which is RF protective clothing.
Incidentally, I just noticed that the RF protective shirt has a
maximum specified protective frequency of 1GHz. Too bad PCS cell
phones operate at 1.9Ghz.
Donga
> Tim May wrote:
>
> >(As with a hollow sphere or any other shape, such as an asteroid with a
> >hollow including the center of mass, there will be no felt
> >gravitational field in the hollow. This applies to objects as big as
> >the earth: one would be weightless through a hollow sphere centered on
> >the center of mass. Because the only "field" one feels is related to
> >the mass _inside the inner sphere_. All contributions from the various
> >bits of the earth above, to the side, on the other side of the hollow,
> >etc., all cancel out exactly (a useful integral calculus exercise to do
> >at least once, just to see how the inverse square law neatly cancels
> >out the contributions).
>
> I don't know of any way to conveniently create a discontinuity in a
> gravitational field or use gravity to model the RF shielding shirt.
That sure is a nice way of telling Tim that he is full of shit. There
*will* be a gravitational force inside of the hollow. The only place
where no gravitational force will be felt is at the gravitational
centroid of the sphere assuming it is in a gravitational Faraday's cage.
There may be a few cases where two or more gravitational centroids
exist, but that is beyound this discussion.
-Donga
Tim May
wrote:
> Jeff Liebermann wrote:
> >
> > I don't know of any way to conveniently create a discontinuity in a
> > gravitational field or use gravity to model the RF shielding shirt.
>
> That sure is a nice way of telling Tim that he is full of shit. There
> *will* be a gravitational force inside of the hollow. The only place
> where no gravitational force will be felt is at the gravitational
> centroid of the sphere assuming it is in a gravitational Faraday's cage.
>
> There may be a few cases where two or more gravitational centroids
> exist, but that is beyound this discussion.
You are incorrect. For the reasons I explained in detail in that last
message, and in some similar messages some months ago (part of the
ongoing issue of people who don't understand these things).
There is no gravitational force inside a hollow sphere inside the
earth, anywhere inside that sphere. The "centroid" of that sphere is
irrelevant.
This can be seen in many ways, from the formal (Stokes's Theorem,
Gauss's Theorem, Shell Law) to the "intuitive" (patches of force from
various parts of the shells outside one's position "cancel out,"
because of the way inverse square laws and solid angles work).
As I said, this is not saying that there is no gravitational force
inside a hole in the ground (e.g., a basement), but a hollow sphere or
other shape (_any_ shape) such that there is no mass inside a sphere
"below" the point in question.
For example, if the earth were to have a _cubical_ void of 1000 miles
by 1000 miles by 1000 miles, at roughly the center of the earth, then
once one got below the "corners" and into the region where no mass was
between oneself and the the center of the earth, there would be no felt
gravitational field (no acceleration).
This is plain old physics. Anyone confused about it should not have
passed first-year physics, or any classes in EE or whatever that
depended on it.
Many texts and online sources explain the math. Here are just two:
<http://geophysics.ou.edu/principles/earth_figure_gravity/index.html>
<http://en.wikipedia.org/wiki/Divergence_theorem>
which summarizes things this way:
"Electrostatics
Applied to an electrostatic field we get Gauss's law: the divergence is
a constant times the volume charge density.
[edit]
Gravity
Applied to a gravitational field we get that the surface integral is
-4嘮 times the mass inside, regardless of how the mass is distributed,
and regardless of any masses outside.
[edit]
Spherically symmetric mass distribution
In the case of a spherically symmetric mass distribution we can
conclude from this that the field strength at a distance r from the
center is inward with a magnitude of G/r? times the total mass at a
smaller distance, regardless of any masses at a larger distance.
For example, a hollow sphere does not produce any gravity inside. The
gravitational field inside is the same as if the hollow sphere were not
there (i.e. the field is that of any masses inside and outside the
sphere only).
See also the shell theorem."
Tim May
Liebermann <je...@comix.santa-cruz.ca.us> wrote:
> Tim May <tim...@removethis.got.net> hath wroth:
> >(As with a hollow sphere or any other shape, such as an asteroid with a
> >hollow including the center of mass, there will be no felt
> >gravitational field in the hollow. This applies to objects as big as
> >the earth: one would be weightless through a hollow sphere centered on
> >the center of mass. Because the only "field" one feels is related to
> >the mass _inside the inner sphere_. All contributions from the various
> >bits of the earth above, to the side, on the other side of the hollow,
> >etc., all cancel out exactly (a useful integral calculus exercise to do
> >at least once, just to see how the inverse square law neatly cancels
> >out the contributions).
>
> I don't know of any way to conveniently create a discontinuity in a
> gravitational field or use gravity to model the RF shielding shirt.
First, I was emphasizing the "field inside is zero" point, coming from
the usual vector calculus stuff that leads to Gauss's Theorem (with
other names like Poisson, Laplace, Stokes usually involved).
This is the point of continuing confusion about how inverse square law
forces work, shown in the most recent message by "Donga."
Second, in this shirt example you keep coming back to, I doubt
seriously that a "conducting cylinder with a vertical slot"
(approximating what this unzipped or unbuttoned shirt or tunic looks
like) acts as an antenna that creates any kind of field across the gap.
Why? Because the conducting cylinder (the shirt in question) is
continuous. No field between one side of the gap and the other (or else
electrons would adjust themselves, and thus no field). Also calcuable
with the surface and volume integrals, but Green's Theorem calculations
get complicated.
> >Silver or "mirror-like" are OPTICAL WAVELENGTH issues.
>
> Yeah, but they're "cool" looking.
Whatever. The point is that you said something about how if Stealth
technology relied on reflection, it would be silvering. This shows a
fundamental misapprehension of optical wavelength reflectivity and RF
wavelength reflectivity.
>
> Well, yes I did intentionally. I don't like quoting large blobs of
> text. I was only interested in mentioning that how the absorptive
> coating works, and perhaps speculating on whether it could be used to
> RF proof clothing. Probably not, because it would be rather heavy.
Indeed, it would be heavy. And a thin conductor (the infamous "tin
foil") would work AS WELL. Do the experiment. Put a radio inside an
aluminum foil enclosure. Listen to the sound (easily heard through
aluminum foil) as the foil is gradually closed up completely.
e
> >
> >In fact, looking at such a liquid metal person, nearly all of the
> >"scenery" would be of a distored, reflected panorama of the viewers
> >half of the full 4-pi solid angle. Do the Gedankenexperiment to see
> >this.
>
> The "thank you experiment"? Huh?
Use the Net, Luke! 10 seconds with Google. Or go straight to the
Wikipedia.
I learned the term Gedankenexperiment in junior high...it was what the
pioneers of quantum mechanics and relativity did to explicate what was
important and what was not.
And, no, the root is not "thank you."
> >This says right off the bat that _most_ of the reflected or scattered
> >illumination (from an optical source in this thought experiment, or
> >from radar or RF in your "RF shielding clothing" example) IS IN FACT
> >REFLECTED OR SCATTERED AWAY.
>
> T2 didn't have an open slot down the front.
Which has nothing whatsoever to do with the specular reflection issue
(pace your point that a reflecting suit would "almost double" exposure
to the head, a claim I showed was clearly false).
You're now conflating your "slotted radiator" (down the front of the
shirt, no less, essentially invisible insofar as the head goes!) point
with your "doubled exposure from reflections" notion.
>
> Similarly, you seem to be lacking in the practical application of some
> of that basic physics. You're also having trouble staying on the
> current subject, which is RF protective clothing.
My T2 example IS about the subject I was discussing, your point that
reflection from the suit will almost double the exposure to the head.
In this case, that of reflection, an optical example IS the same as an
RF example (simple ray-tracing, to first order).
So if the head of a guy wearing a T2-type surface material (reflective)
is not doubled in apparent brightness--which it will not be, for the
various reasons I gave about solid angles and all--then the same
applies with millimeter or centimeter waves.
(With larger waves, diffraction effects become more significant. Not
that this will increase the exposure to the head, either, for obvious
reasons. In fact, just the opposite.)
You need to study up on Maxwell's equations. Not the gory double
integral and triple integral "mechanics," from a calculating point of
view, but from the conceptual point of view: fluxes, divergences,
surface integrals (conceptually), volume integrals (conceptually),
curl, etc.
Yours and Donga's claims about fields inside shells or spheres are
simply wrong-headed.
--Tim May
Tim May
Another point:
In article <060420060950279058%tim...@removethis.got.net>, Tim May
<tim...@removethis.got.net> wrote:
> > Well, yes I did intentionally. I don't like quoting large blobs of
> > text. I was only interested in mentioning that how the absorptive
> > coating works, and perhaps speculating on whether it could be used to
> > RF proof clothing. Probably not, because it would be rather heavy.
>
> Indeed, it would be heavy. And a thin conductor (the infamous "tin
> foil") would work AS WELL. Do the experiment. Put a radio inside an
> aluminum foil enclosure. Listen to the sound (easily heard through
> aluminum foil) as the foil is gradually closed up completely.
Something I should have brought out more clearly is this:
Jeff is confusing, or conflating, two different issues:
1. The electric fields INSIDE a material, in this case the
RF-protective shirt (aka "tin foil hat" in generic terms).
2. The returned signal from a radar or other RF source that is
illuminating an aircraft.
In the first case, all that matters is that no electric field be inside
the conductor (or at least that it be the appropriate number of
decibels down).
In the second case, even a reflective surface has a return signal of
some size, the familiar radar cross-section (RCS). Even a
superconducting hollow cube--which I'm sure even Jeff will agree has no
E-field inside the cube!--will have an RCS which depends on the
orientation with respect to the source. If a facet of the cube is
orthogonal to the source, a maximal RCS. If not orthogonal, the RF
reflects away, with various second- and higher-order diffraction
effects possibly contributing to a small RCS.
Stealth technology is not about eliminating E-fields inside the
aircraft (that is TEMPEST technology). Rather, it is about minimizing
RCS. So for its intended use, a combination of reflection (the faceted
planes of the surface of the F-117A, for example) and absorbtion (the
ferrites and internal reflecting surface baffles) are used.
For the RF-protecting shirt, RCS obviously DOES NOT MATTER.
This is why the calculations for Faraday cages are so applicable.
And for this use, a conductive material of even aluminum foil thickness
works perfectly well.
All of the stuff about ferrite beads and other RF-absorbing material is
missing the point.
--Tim May
Donga
> Donga wrote:
>
> > Jeff Liebermann wrote:
> >
> > > I don't know of any way to conveniently create a discontinuity in a
> > > gravitational field or use gravity to model the RF shielding shirt.
> >
> > That sure is a nice way of telling Tim that he is full of shit. There
> > *will* be a gravitational force inside of the hollow. The only place
> > where no gravitational force will be felt is at the gravitational
> > centroid of the sphere assuming it is in a gravitational Faraday's cage.
> >
> > There may be a few cases where two or more gravitational centroids
> > exist, but that is beyound this discussion.
>
> You are incorrect. For the reasons I explained in detail in that last
> message, and in some similar messages some months ago (part of the
> ongoing issue of people who don't understand these things).
>
> There is no gravitational force inside a hollow sphere inside the
> earth, anywhere inside that sphere.
You use the word "earth" (sic), but ignore the gravitational forces from
the sun, the earth, and from everything else that exists.
I have performed the calculations required to derive your conclusion.
But, I am an engineer with practical experience and I disagree with your
theoretical conclusions.
I used to have an old basketball with a loose piece of the valve stem
inside. There is your hollow sphere. That loose piece acted as if it
were affected by gravity.
Oh, you want to talk about the impossiblility where only your objects of
study exist? You know, a place doesn't exist...
Even if you could find a practical use for your hollow sphere inside a
gravitational Farady cage, gravitational forces would still produce an
excelleration on any mass contained within this hollow sphere.
And don't tell me that you weren't thinking about keeping the hollow
sphere void of any mass. What good what that be? As you said, the
gravitational forces within a hollow shpere are the same as those within
a "solid" (massive) shere. If that's what you're talking about, then
its just so much mental banging.
And even if you could find a use for a hollow sphere without a mass
inside, then you still will never experience your Utopian zero net
gravity. You see reality has this concept of ridgidity where even the
most ridig object still is not a theoretical solid.
What this means for the current discussion is that your perfect hollow
sphere will not be able to hold an exact shape. Therefore, a
gravitational force will always be present at any point in space (except
a few special cases). Your claim only holds for perfect theoretical
cases.
If the Earth were hollow, and a human could get inside there, then the
only gravitational force they would experience would be that of the sun
and moon with minor forces from other objects. But, there would still
be gravity. It would change over the year much like our tides.
-Donga
John Mann
Donga wrote:
> You use the word "earth" (sic), but ignore the gravitational forces from
> the sun, the earth, and from everything else that exists.
>
> I have performed the calculations required to derive your conclusion.
No you haven't; you're doing what I occasionally do; using big, made-up
words to convince people I know something on a subject when I don't
know shit. (What in *fuck* is a "centroid"? Even *I* wouldn't be so
blatantly clueless as to make up a word like that in a roomful of
phd's.)
Give it up, weak sister.
Donga
Note the use of the word "gravitational" in respect to a centroid:
http://en.wikipedia.org/wiki/Centroid
-Donga
John Mann
Donga wrote:
> John Mann wrote:
> > No you haven't; you're doing what I occasionally do; using big, made-up
> > words to convince people I know something on a subject when I don't
> > know shit. (What in *fuck* is a "centroid"? Even *I* wouldn't be so
> > blatantly clueless as to make up a word like that in a roomful of
> > phd's.)
> Note the use of the word "gravitational" in respect to a centroid:
> http://en.wikipedia.org/wiki/Centroid
You *still* did not use the word correctly, now that we've proven it
exists.
Jeff Liebermann
>What in *fuck* is a "centroid"?
A centroid is 100 hemorrhoids. Probably itches rather severely. I'll
pass on the obvious one liner as this is a family newsgroup.
Donga
We did not prove anything. I simply showed you that a minute on google
looking up the word "centroid" would have saved you from looking LIKE A
STUPID FUCKING IDIOT, AGAIN!!! FUCK YOU, YOU FUCKING FUCKHEADED FUCK!!!
Since you like to tell people how much you read books, why didn't you
look up the word in a dictionary?
If you can create a post worth reading, explain what was wrong with my
use of the word "centroid".
-Donga
nikhil sharma
Explore more on https://sleepgift.ca/
Julian Macassey
On Sat, 5 Aug 2023 23:00:28 -0700 (PDT), nikhil sharma
<nikhils...@gmail.com> wrote:
> We are thrilled to share an incredible milestone our team at
> SleepGift achieved — validation and proof of the EMF shielding
> feature in our weighted blankets made with nano-sized silver
> mineral infusion in the fabric. Through the rigorous testing
> processes by (Ontario Tech University) specializing in EMF
> detection methods. They have validated the effectiveness of
> integrating silver into each blanket fiber, which effectively
> blocks out potentially harmful wave
Subject "for the tin hat folks"
Peddling junk science for tin hat folks.
I hope he dies in prison.
If you need a laugh...
--
The NHS will last as long as there are folk left with faith to
fight for it. - Aneurin Bevan
John
> On Sat, 5 Aug 2023 23:00:28 -0700 (PDT), nikhil sharma
> <nikhils...@gmail.com> wrote:
>
>> We are thrilled to share an incredible milestone our team at
>> SleepGift achieved — validation and proof of the EMF shielding
>> feature in our weighted blankets made with nano-sized silver
>> mineral infusion in the fabric. Through the rigorous testing
>> processes by (Ontario Tech University) specializing in EMF
>> detection methods. They have validated the effectiveness of
>> integrating silver into each blanket fiber, which effectively
>> blocks out potentially harmful wave
>
> Ironically the grifter who posted the above ignotred the
> Subject "for the tin hat folks"
>
> Peddling junk science for tin hat folks.
>
> I hope he dies in prison.
>
>> Explore more on https://sleepgift.ca/
>
> If you need a laugh...
favorite materials of magical mineral health kooks. Personally I would
*love* to have silver nanoparticles infused into my blanket; I assume
they'll never come free and get infused into my lungs instead!
john
pH
effective against RF and coronal mass ejections?
It's been pretty okay against orc blades and Nazgul spears,though.
pH in Aptos
> https://sleepgift.ca/
Jeff Liebermann
wrote:
>Is Jeff insinuating that my Mithril body armor may also be less than
>effective against RF and coronal mass ejections?
coronal mass ejections because I don't smoke Corona Cigars.
<https://www.coronacigar.com>
>It's been pretty okay against orc blades and Nazgul spears,though.
>pH in Aptos
--
Jeff Liebermann je...@cruzio.com
PO Box 272 http://www.LearnByDestroying.com
Ben Lomond CA 95005-0272
Skype: JeffLiebermann AE6KS 831-336-2558
BCFD36
in a high wind.
--
Dave Scruggs
Captain, Boulder Creek Fire (Retired)
Sr. Software Engineer (Retired, mostly)