Re: nobody knows whos maggie at fla teslathon?

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Gerry Reynolds" <gerryreynolds-at-earthlink-dot-net> 

I think the P in the equation is the true power from the line (ie VA * PF).
Seems like this result is good.  Working backwards,  16 feet would
correspond to a Power of 12.7 KWatts.  With a 18 KVA input that would imply
a power factor of .7   I don't yet know what is possible with a SRSG, but I
believe the best PF one can get with a static gap is something around 0.5
without allowing resonant rise to overvolt the transformer (per Richie
Burnett).

Others please comment.  This is an area I want to learn more about.

Gerry R
Ft Collins, CO

 > Original poster: "RMC" <RMC-at-richardcraven.plus-dot-com>
 >
 >  > It doesn't take a lot of time to realise the magnifier in question
isn't
 >  > much of a very efficient device.
 >  > Power input 18 kVA and observed max arc lenght of 16 feet.
 >  > Formula ,frequently mentioned on this list,L(inches)=1.7*SQRT(P) gives
19
 >  > footers for that power.
 >
 > That is still 84% of what the empirical formula estimates. Besides, the
 > empirical formula I think has been developed to quantify two-coil TCs, not
 > magnifiers.
 >
 > I think to say that the device "isn't ...very efficient" is a bit unfair.
 > How much better have other people achieved...or have you achieved?
 >
 >
 > RMC, England
 >
 >