Re: Non-Radiative Evanescent Waves are back in the news, Political (fwd)

["Tesla list" <tesla@xxxxxxxxxx>]

---------- Forwarded message ----------
Date: Sun, 10 Jun 2007 17:29:26 -0700
From: Ed Phillips <evp@xxxxxxxxxxx>
To: Tesla list <tesla@xxxxxxxxxx>
Subject: Re: Non-Radiative Evanescent Waves are back in the news,
      Political (fwd)

Tesla list wrote:

>---------- Forwarded message ----------
>Date: Sun, 10 Jun 2007 14:51:33 -0700
>From: Jim Lux <jimlux@xxxxxxxxxxxxx>
>To: Tesla list <tesla@xxxxxxxxxx>, tesla@xxxxxxxxxx
>Subject: Re: Non-Radiative Evanescent Waves are back in the news,
>      Political (fwd)
>
>At 11:13 AM 6/10/2007, Tesla list wrote:
>
>  
>
>>----
>>
>>        For the record I ran a coupling calculation for the coils 
>>in their experiment and get a value of ~0.0034 [if I didn't miss a 
>>decimal point] so even a Q of 300 at each end would give very tight 
>>coupling.  No miracle, no breakthrough, no nothing.
>>
>>        When I get a chance I'll try to estimate the radiation from 
>>the transmitter.  Suspect it will be significant.
>>
>>Ed
>>    
>>
>
>I'd compare the radiation resistance of the transmitter coil to an 
>estimate of the resistance (from Q?)
>
>The radiation resistance will tell you how much energy gets out into 
>the far field.  In theory, the ratio of radiation resistance to loss 
>resistance will tell you how much gets radiated vs lost as heat. And, 
>the Q tells you how much energy is in the field vs how much is lost.
>  
>
    The radiation resistance works out to about 0.0165 ohms - that's the 
easy part.  I made a calculation based on an estimated loaded Q of 300 
and 150 watts total power dissipated (most in loss resistance of course) 
and got about a half watt as I posted earlier.  If I use an approximate 
Q of 800 of course I get a larger number, around a watt and a half.  
Gotta think about it.  In either event a significant amount of power as 
far as interence generation is concerned and obviously unacceptable, 
even in the ISM band.

Ed

Ed