Re: coupling losses ? (fwd)

["Tesla list" <tesla@xxxxxxxxxx>]

---------- Forwarded message ----------
Date: Fri, 21 Sep 2007 10:01:36 -0400
From: Dave Pierson <davep@xxxxxxxx>
To: Tesla list <tesla@xxxxxxxxxx>
Subject: Re: coupling losses ? (fwd)

 
>If you start at 10KV on the primary, and the secondary is maybe 2" away, 
>there is some distance there and coupling will be lower.
     Only a tiny amount.

> As there is some distance, the 10KV line
     'line'?

> has to drop over the square of the distance ?
    'inverse square law' commonly oversimplified.
    Applies to 'radiation' (which is not the case here)
    and, roughly, applies in 'far field' (say 1/2 wavelength?) out:
    say 100s meters in Tesla Coil Case.

>Saying so,
    cf as above.

> if starting out with 50% coupling,
     We're not, as above.

> then it would be more like only 5KV
     The coupling loss (moniro) affect energy transfer, rather than
     voltage.

> actually gets onto the secondary coil?
     I wouldn't say so.

>After some of my own analysis of sparks gaps, its probable that my 10KV source
> will not turn off over the spark gap until about 8kv.
     Need to know how the coil system is wired.  Gap losses are low, once
     on.

> So if we take 8KV as the actual primary voltage, 
>loose 50% due to low coupling, then this leaves us with only 4KV actually 
>driving the secondary coil?
     cf as above.

>On this basis the voltage gain of the secondary would have to be 50 to gain 
>200KV output? Does the "50" not mean the "Q" of the coil?
     Turns ratio, primary to secondary, is a bigger influence.

>or what Tesla called the "magnifying factor" ?
     I disrecall how he used that term.
     best
      dwp