TSSP: List Archives

From: boris petkovic
Date: Sat, 23 Dec 2000 12:01:55 -0800 (PST)
Subject: Re: [TSSP] Final solution...great chance for it

Hi Paul,

> 
> > > > -----
> > I guessed you would say that.
> > What about (computing time), simulating 100 turn
> coil
> > response when step function applied ,say at 
> pentium
> > processor?
> > -----
> 
> Funny you should say that. I have a time domain
> simulator
> which just happens to have 100 elements...its a
> while since
> I've used it, and it doesn't do the longitudinal
> coupling
> stuff. I don't have any timings for it, but its
> *very* slow,
> I seem to recall an hour or more to get a solution
> for a
> transient.
----
I see..
---

> 

.
> 
> > Physical background of the process is better
> explained
> > in his case.
> 
> Understood. I'd like to see more - will look forward
> to
> the post arriving.
---
Next week I will send you the package.I tried
today,but I didn't make it .Post offices downhere in
Croatia were closed (Holidays..I didn't count on
that).
Do you have The Corums' paper on their production of
fire balls.?If you don't have it I'd like to send it .
Althought I completely disagree with their resonator
theory,the paper of production of fire balls is a must
see.
--- 
   
> > Once,you re intersted in predicting Q of
> secondaries
> > ,right?
> > Now,I will ask you quite a diffucult question.
> > What do you think could be the highest possible Q
> of
> > the structure like Tesla's 100 turn 100"*100''
> extra
> > coil from CS?
> > Imagine  there is perfect ,loseless
> isolation,super
> > conducting winding and enviroment,and the only
> > losess comes becouse of radiating.
> > I don't expect concrete number ,but try to guess
> the
> > ballpark.
> > ------
> 
> OK, the only loss is due to radiation. I worked this
> out
> once, a long time ago. Can't find my figures so I'll
> do
> it again...
> 
> Consider the lossless TC resonating in free space
> with angular
> freq w and stored energy E. The Q factor is then
> 
>  Q = wE/P
> 
> where P is the radiated power. From a sufficient
> distance away
> it resembles a Herztian dipole with a dipole moment
> qh where
> q is the displaced charge and h the coil length. 
> Since V = sqrt(2E/Cee) then q = CesV = Ces
> sqrt(2E/Cee).
> 
> The total power radiated from a Herztian dipole is
> obtained
> by integrating the Poynting power density over the
> surface
> of a large bounding sphere. I'll spare the details
> and say
> 
> P = (1/12pi) * z * w^4 * q^2 * h^2 / c^2
>   = (1/6pi) * z * w^4 * Ces^2 * h^2 * E/Cee/c^2 
> Watts
> 
> where c = 300 * 10^6 metre/sec, z = 377 ohms.
>  
> Thus 
> 
> Q = wE/P = 6pi * c^2 * Cee / (z * w^3 * Ces^2 * h^2)
> 
> Say Cee = Ces = 30pF;  w = 2.pi.10^5 rad/sec;  h = 1
> metre.
> 
> Then Q is approx 6 * 10^8, pretty high.
> 
> Now is that in the right ballpark? Are we safe to
> ignore 
> radiation? I get the feeling you're going to tell me
> I'm
> missing something :)
----
I will not comment  calcs at all,neither I will
explain my figures bellow.
I will just offer 'my' ballpark:
Q=1000...10000
----

Regards,
Boris



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Maintainer Paul Nicholson, paul@abelian.demon.co.uk.