3 ms·

I found some napkin math on stackexchange: http://electronics.stackexchange.com/questions/4664/stealing-energy-from-radio-towers-or-power-lines http://electron

by gibybo 13y ago

I found some napkin math on stackexchange: http://electronics.stackexchange.com/questions/4664/stealing-energy-from-radio-towers-or-power-lines http://electronics.stackexchange.com/questions/4664/stealing... One of the answers has the following math: For a large radio tower (100kW), you'd get roughly 80 W per square meter of receiving antenna area at 100m distance. Since the typical distance to a nearby radio tower is probably more like 5 miles, we can reduce that to ~12mW for a 1m^2 antenna. For the size of a phone, we're talking more like 0.10mW, which is about 10,000 times less than a typical phone needs (assuming it was charging 24/7 with 100% efficiency). Although, the answer gives the formula as 100kW/(4 * pi * r^2), which seems to give 0.80 W/m^2 instead of 80 W/m^2, so perhaps divide my other answers by 100 (and make that 1 million times lower than the power necessary to power a phone) EDIT: I previously stated the formula gave the wrong units which was a mistake.

leoedin 13y ago
I think the equation given does give units in watts/m^2? Unless I'm missing something, power has the units of W, and the only dimensions on the bottom of the equation are the radius, which is meters. Because it's r^2, the result is in W/m^2. For a non-directional beam, it would be P/(4.pi.r^2).
- gibybo 13y ago
  Yea my mistake, I missed the units on the radius :( thanks!
VLM 13y ago
As an off the cuff engineering estimate an AM crystal radio is pretty efficient internally and just barely converts a multi kilowatt transmitter output into barely audible sound inside an earphone, given a substantial antenna and grounding system. So for multiple closer antennas, and something much more electrically demanding than a crystal radio, like a smartphone, its believable. The biggest problem is going to be PR. So wifi milliwatts of microwave power tens of feet away make hypochondriacs and conspiracy theorists think the government is controlling their thoughts using radio waves... I'm guessing "tens of watts in a ISM diathermy band" is going to substantially encourage them.
mturmon 13y ago
It just occurred to me that the / (4 pi r^2) factor may not be relevant here. The reason is that the acoustic wave could be steered (http://en.wikipedia.org/wiki/Beam_steering http://en.wikipedia.org/wiki/Beam_steering) so it does not propagate evenly over all the solid angle. Perhaps the steering could be done cooperatively with the device(s) being charged. I.e., devices announce their location to the charger, and the charger steers the beam towards them. Maybe that would bring the power required down to a reasonable range.

by gibybo 13y ago

leoedin 13y ago

I think the equation given does give units in watts/m^2? Unless I'm missing something, power has the units of W, and the only dimensions on the bottom of the equation are the radius, which is meters. Because it's r^2, the result is in W/m^2. For a non-directional beam, it would be P/(4.pi.r^2).

gibybo 13y ago
Yea my mistake, I missed the units on the radius :( thanks!

VLM 13y ago

As an off the cuff engineering estimate an AM crystal radio is pretty efficient internally and just barely converts a multi kilowatt transmitter output into barely audible sound inside an earphone, given a substantial antenna and grounding system. So for multiple closer antennas, and something much more electrically demanding than a crystal radio, like a smartphone, its believable. The biggest problem is going to be PR. So wifi milliwatts of microwave power tens of feet away make hypochondriacs and conspiracy theorists think the government is controlling their thoughts using radio waves... I'm guessing "tens of watts in a ISM diathermy band" is going to substantially encourage them.

mturmon 13y ago

It just occurred to me that the / (4 pi r^2) factor may not be relevant here. The reason is that the acoustic wave could be steered (http://en.wikipedia.org/wiki/Beam_steering http://en.wikipedia.org/wiki/Beam_steering) so it does not propagate evenly over all the solid angle. Perhaps the steering could be done cooperatively with the device(s) being charged. I.e., devices announce their location to the charger, and the charger steers the beam towards them. Maybe that would bring the power required down to a reasonable range.