How much EM energy, of a type that could be absorbed by a typical metal antenna, is there in an average spot? Is it significant? Less than a watt?
The EM spectrum is huge and you’d have to define a range to fully address your question. For example, you’re probably referring to the VHF/UHF range of spectrum (say from 30MHZ-2GHz) but sunlight, being a type of EM emission, is also readily absorbed by the antenna, too.
What you are probably after is what is referred to as the RF noise floor. You can define a range (ie VHF/UHF). This is usually measured in dBm and varies quite a bit based on the RF environment. A very urban, downtown office building with a pager tower on the roof is clearly going to be much noiser than the middle of the Siberian tundra. That said, an absorbed watt is not likely to be approached accidentally.
I thought that I was defeining enough of a range when I restricted it to those frequencies that can be absorbed by a typical metal antenna.
What do you mean by saying that visible light can be absorbed by a metal antenna? I do not care if the produced energy is only heat, I am looking for measurable electrical potential.
What I was getting at is this. A radio is little more than an overglorified transformer, absorbing the EM radiowaves, amplifying the potential created, then producing sound based on the patterns encoded within the EM radiation. What I want to know is how much energy is there in a typical city setting (but the ammount in a secluded area, or even the dark side of the moon, would also be interesting) for all of the EM that can be recieved by a typical antenna and be converted into electrical work?
It’s no different from asking a photopher how bright is the light where I am? Well, where are you? If you are standing in direct sunlight with no clouds I can make an accurate estimate based on it being 93 million miles to the sun. If you are somewhere else get a light meter. and measure it yourself.
What about at a place like the dark side of the moon? Wouldn’t the universal background EM be fairly constant for places that are not near a localised source?
Ok, now the question shifts to your definition of ‘typical antenna’. For example, a wire whip antenna clipped to a certain length is now tuned to frequencies with corresponding wavelengths (as well as 1/2, 1/4, 1/8, 3/4, etc). If you hook that wire whip to a ‘perfect receiver’ (it’s in the cabinet, there next to the frictionless pulley and the ideal capacitor), you’ll find little noise peaks at these tuned points. Using complex geometry to try to design wideband antennas that scoop up signals over a large band of spectrum or are tuned to specific bands with usable gain remains an ongoing problem for antenna designers. Antennas like discones and yagis are familiar but have tradeoffs.
The television beam on a typical home, for example, must be pointed roughly in the right direction (compass and elevation) to work right. Additionally, if you were to rotate the antenna axially 90 degrees, you’d find quite a bit of interference. Polarization matters!
I’m not trying to be a stick in the mud. This is a complex question with more factors than meets the eye. That said, I did locate a study from New Zealand addressing this question:
(.pdf!) http://www.med.govt.nz/rsm/publications/pibs/pib32.pdf
Page 21 shows a handy chart. In summary, the noise floor they report is roughly between -100 and -125dBm. This is for a band from about 50 to 250 MHz. Appendix 2 has more detail.
I thought that near perfect antennas have already been developed. I remember reading an article in Scientific American some years ago about how engineers had discovered that antennas shaped as fractals received all frequencies equally.
You can’t provide a guesstimate as to a reasonable range of watts that could be harnessed?
[QUOTE=Muad’Dib]
What about at a place like the dark side of the moon? Wouldn’t the universal background EM be fairly constant for places that are not near a localised source?[/QUOTEYes, now you just have to describe a typical antenna. A “typical” antenna may or may not see the right bandwidth for background microwave radiation. Size of the antenna makes a huge difference too. Does it have a collector dish to gather over a wider area?
Your question doesn’t have a single, definite answer. Maybe this will give you a feel for the about of EM in the radio frequency spectrum.
If there are 4 10 kW transmitters using quarter wave vertical tower antennas 50 miles from you the EM power from them at your location is about 2 microWatts/m[sup]2[/sup].
The problem is we don’t really know how many radio frequency transmitters are in the vicinity of the “average spot.” I just chose 4 off the top of my head. In rural “average spots” there probably wouldn’t be that many. In a city there might be a couple of 50 kW transmitters only a few miles away.
I see what you mean.
I will work on this some more.
So, are those few microwatts enough power to do anything useful, like run a wireless nightlight?
Well they are enough to be detected by a radio receiver and so provide cheap entertainment for the masses. 
There’s entertainment in radio waves, you say? And here I was thinking that music came from the internet…
I meant, is there enough power to do something useful without the need for a separate power source, but now I see there are crystal radio kits that don’t need batteries.
True enough. I haven’t looked into the specifics of crystal receivers but they probably wouldn’t produce a useable about on so little power.
What about general background EM radiation? Forgiving some localized energy source, isn’t that fairly constant despite your location? I know that a typical radio can tune into a wide range of frequencies, what is the limit to this range and how much power could be harnesed from the that segment of the EM spectrum?
Here’s a little on EM that gets through from space:
http://linas.org/theory/seti.html
Wierd units though; 1E-20 Watts / (meter[sup]2[/sup] steradian Hz). I’m not sure how you’d sum that to cover the entire sky, and the full bandwidth.
Not so weird, as that’s what EM radiance is: energy per unit time per unit of projected area per unit solid angle per unit frequency.
In general we’d have to know the incoming radiance as a function of direction and frequency, and evaluate a double integral over both of those variables. If you gave me nothing else though, I would just multiply that quantity above by one hemisphere (2 pi steradians) and the size of the receiver’s bandwidth in Hertz. That product would give you a good order-of-magnitude answer — accurate enough to satisfy an astronomer, at least.
Too techy for me to follow, but it’s an interesting question.
Variation one: How much electromagnetic energy am I absorbing per hour, sitting here? In watts? Order of magnitude is plenty accurate enough, you could paint with a pretty broad brush as far as I’m concerned. I’d bet it is less than a watt.
Variation two: If I stick a ten-foot metal pole out the window, and put a rectifier circuit on this end to convert it to DC or 120 Hz, what do I get in watts?