I can't bear to watch. I have a heavy load to bear. There is a bearing on my axle. Is that animal a bear? Does the use of bear have a bearing on language?

 

I can't bear to watch. I have a heavy load to bear. There is a bearing on my axle. Is that animal a bear? Does the use of bear have a bearing on language?

I cannot bear to bear that bear.
—Mr. Ranger

 

RE:""Will metal block all frequencies of radio waves, or are there some frequencies metals won't block?""

Firstly, should analyze separate the mag-field and separate E-field. Only when both are kept well diminished, the insulation is complete. So, magnetic field may be good attenuated in the ferromagnetics, as better the mju, as better, BUT all the ferromagnetic materials contains the grains thus the frequency giving wavelength above the grain size means unefficiency of subsequent ferrite for so high frequency. Most strongest material used in experimental physics for this aim is mju-metal, and it`s only ill point is it price about 60-fold gold price. So, my experience is that mag field be best ecranated with multi-layer sandwich, containing ferromagnetic layer, insulator, antimagnetic layer (aluminium(, film again, then second steel +paper+plus aluminium plus paper and so on about 6-8 layers giving attenuation about 100x for DC and few millions for moderate RF. Never measured this over 3.5 GHz. Ah, ya, ferromagnetic film must have drilled hole to hole like loudspeaker screens if steel is 0.5 mm thick, then drillings are 0.5 mm as well or 0.7 and crisscrossing pattern let be impossible to draw straight line between them - this measure makes that mag field is not concentrating in ecrane end cuts but makes a thousands of poles in each drilling.

When yet the clear RF is attenuated in the metall plate, there takes a place Focault effect having dependency - swirl radius aka Focault deepth vs frequency depending on resistance, mju and epsilon of material. Well may read about it at https://en.wikipedia.org/wiki/Skin_effect stating that delta=sqrt(ro/w/mju) where w=2pi*f if f<<1/ro/epsilon. If any swirl may "go through" the thickness of platelet, means ecranation is uneffective.

So, now about E-field - there the completely "transparent" are materials having no molecular resonances in the range. If resonance is rather near, then it is semi-transparent or terminologically more correct to say, lossy. Lossy dielectrics makes loss factor be proportional to square root of frequency. If for small loss factors being far beyound the 10 grades between active and reactive currents sin(a)~~(a) in radians, making as the consequence sin(a)~tan(a)~cos((90-a)=cos(fi) or other words Loss Factor~Power Factor. Thus the thermal heat flux in lossy capacitor is P(th)=P(react)*tan(delta) where tan(delta needed freq)=tan(delta given freq)*sqrt(needed freq/given freq).

Yet while the metalls have generally one resonant or worst case two, the dielectrics may have a multiple, the molecular rotational, vibrational etc resonances standing between 1-10 GHz for macro-molecular insulators and 10-1000 GHz for short-Dalton plastics.

So, I would guess the any good conductor metall like aluminium may work wery well as the ElMag insulator for RF, however the multiple-layer pie is the absolute best.

P.S. If inside the ecrane stays coil, or even worser, the resonant coil, the empyrical thumb-rule must be observed- best case the ecrane diameter is 3*coil diameter or most ever worst case scenario, the 1.5 diameter to suppress too much inquisitive loss.

 

So if I had a powerful enough transmitter and fired a focused beam of radio waves at 2.8 gigahertz at a. Solid Wall made of mild steel, and not ground, the radio waves would pass through the wall and a receiver on the other side would detect the signal?

 

RE: Man10
Yes, even if the signal is not so much powerful and even if the metal walls means a full Faraday Cage and even if that metal is well earthened.
Okay, shall give an example: 50 mW 868 MHz packet transmitter LORA(TM) protocol inside the 2 mm thick air-tight pressurized sea-container of zinced mild steel earthened within 4 Ohms, to identical receiver outside the container - 3 meters afar gave 20% of packet loss factor, means "telegram" was re-sent second time to become booked as received.

 

 

The OP is essentially asking about EMI shielding without using the term or seeing the connection to EMI shielding (which can include shielding (reflecting) and absorbing (converting the signal to heat energy via eddy currents in the metal)).

 

So if I had a powerful enough transmitter and fired a focused beam of radio waves at 2.8 gigahertz at a. Solid Wall made of mild steel, and not ground, the radio waves would pass through the wall and a receiver on the other side would detect the signal?

Yes
your waves are attenuated , so still there , but smaller.

Its "just" a mater of detecting,

refer back to Xrays .

when you started this , it was a general question, a though experiment ?

It now seems to be turning into how do I make a ray gun pointing through a wall ,

What are your intentions ?

 

Yes
your waves are attenuated , so still there , but smaller.

Its "just" a mater of detecting,

refer back to Xrays .

when you started this , it was a general question, a though experiment ?

It now seems to be turning into how do I make a ray gun pointing through a wall ,

What are your intentions ?

Just to learn how radios work.

 

Will metal block all frequencies of radio waves, or are there some frequencies metals won't block? For example will lower frequencies, penetrate metals better? Is attenuate a more accurate term than block?

Kind of depends on what you mean by "block" . What your application. What I am thinking about is Cu a conductor but put something in a Cu Faraday cage and ground it will block rf by shunting it to ground, as long as the wave length is longer than any holes in the box.

 

Kind of depends on what you mean by "block" . What your application. What I am thinking about is Cu a conductor but put something in a Cu Faraday cage and ground it will block rf by shunting it to ground, as long as the wave length is longer than any holes in the box.

Your typical Cu Faraday cage doesn't work by shunting current to ground. If that was necessary RF shielding in airplanes wouldn't work. A "perfect" Faraday cage does not block EM radiation but it can attenuate /reduce the in/out going signals to acceptable but not totally undetectable levels.

https://www.google.com/books/editio...gbpv=1&dq=isbn:0780360249&printsec=frontcover

We will assume the emitting source is placed at a reasonable distance from the wall of the enclosure (inside) such that we are concerned with far field radiation (radiation that has escaped the antenna; 1/6 wavelength is often used as an approximate distance for predominance of radiation field in shielding applications) propagating outwards towards the enclosing shield wall.

The attenuation provided by the shield results from three mechanisms: (1) reflection of an electromagnetic wave when it encounters an impedance disontinuity, e.g., the air to metal impedance discontinuity as the wave encounters the shield (2) absorption within the shield material of portions of the wave energy not reflected as it transfers some energy in heating the shield and (3) possible additional reflections within the shield and at the impedance discontinuity as any remnant of the wave encounters another metal (typical shield material) air boundary. The diagram below is for a wave approaching from outside; simply reverse the label, i.e., let "inside of enclosure" in the diagram be "outside of enclosure."

Electromagnetic radiation is primarily shielded by reflection via mobile charge carriers (electrons or holes) which interact with the radiation. High conductivity of the shield is not required, e.g. on the order of 1 ohm is usually sufficient. Electrical conductivity is not the criterion for shielding (though conductivity enhances shielding) though, since that would require connectivity in the conduction path, e.g., to ground.

 

It now seems to be turning into how do I make a ray gun pointing through a wall ,

What are your intentions ?

Sounds like he either wants to blast his ray-gun through a wall or protect himself from someone pointing a ray gun at his wall. So, he's either Daffy or Marvin.