On Sun, 20 Sep 2009 07:56:11 +1000, you wrote:
>In reply to John Fields's message of Sat, 19 Sep 2009 10:08:09 -0500:
>Hi,
>[snip]
>>On Sat, 19 Sep 2009 14:18:54 +0200, you wrote:
>>
>>>John, sorry for the late answer.
>>>
>>>Unwanted induction heating on rings necklaces etc: they say it doesn't
>>>happen because you need very fine tuning to receive (see the TED video
>>>I linked to, the guy walks happily through the power beam, same thing
>>>for the original MIT research team photographed while sitting in the
>>>beam, photo shown in the video)
>>
>>---
>>As far as I know, the ring or necklace would act like a shorted
>>single-turn secondary of a transformer and would heat up without regard
>>to the frequency of the field, the heating depending only on the turn's
>>resistance, its orientation relative to the field it was in, and the
>>intensity of the field.
>
>I agree, however because a metal ring wouldn't be tuned, the energy transfer
>would go as 1/r^2, and be just as inefficient as an air core transformer at
>considerable distance, so I don't think heating would be a problem unless you
>were quite close to the source.
---
I'm pretty sure that, even tuned, power falls off as 1/r² and, because
of the inefficiency, the intensity of the field would have to be quite
high in order to excite the 'secondary' in the device being charged,
thereby also inducing currents which could be quite high in 'shorted
turns' being worn by the user.
What tuning gets you, though, is elimination of the reactive terms in
the receiver's front end, so that all that's left to deal with is the
resistance of the load.
In this case that would be the resistance of the tuned secondary itself,
and the resistance of the charger circuitry.
Just for grins, let's say that a cell phone needs to be able to charge
its battery in one hour, and to do that it needs to, from the field,
extract enough power to put 5 volts RMS across the charger and push 100
milliamperes through the battery.
Further, let's say that the receiver's coil comprises 100 turns of
copper wire.
Ohm's law tells us that the resistance of the load would be:
E 5V
R = --- = ------ = 50 ohms
I 0.1A
and it would dissipate:
P = IE = 0.1A * 5V = 0.5 watts.
Now assume that someone wearing a necklace with a resistance of 0.1 ohms
walks into the field.
Since it's only a single turn the voltage across it will be 1% of the
voltage induced in the receiver's coil, so the current through it will
be:
E 0.05V
I = --- = ------- = 0.5 ampere
R 0.1R
and the power it would dissipate would be:
P = 0.05V * 0.5A = 0.025 watts
That's trivial, so under those conditions it looks like a safe scenario.
---
>IOW best placement in the home for the
>transmitters would probably be in the ceiling.
---
For a small device like a cell phone I think a pad might be more
efficient, but still not as good as a conventional, properly designed
switching power supply with load detection wired to the mains.
---
>BTW while we are on the topic, consider that it might be possible to use the
>lower Van Allen belt as the transmitter, allowing reception of free power.
>
>(The belt itself is of course powered by the solar wind).
>
>There should be a point where the strength of the Earth's magnetic field
>results
>in a cyclotron frequency that has a wavelength long enough to reach the Earth's
>surface, ensuring that we are within 1 wavelength of the "transmitter".
>I have for some time suspected that this concept may lie behind some of the
>stranger free energy devices.
---
I'm not conversant in those areas, so I'll just sign off with: "Perhaps"
;)
