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Re: Why does top capacitance work? (fwd)





---------- Forwarded message ----------
Date: Mon, 24 Feb 1997 20:18:55 -0500 (EST)
From: richard hull <rhull-at-richmond.infi-dot-net>
To: Tesla List <mod1-at-pupman-dot-com>
Subject: Re: Why does top capacitance work? (fwd)

snip


>That is exactly it. Doesn't make much sense from the angles we've 
>been exploring so far does it? With a bit of luck I might just find 
>out why when I have completed the exercise which will still take a 
>couple of weeks. I have over thirty designs to analyse thanks to 
>everyone and there is much calculation needed to extract the 
>quantities which I am examining - in fact the chart I am drawing up 
>is enormous.
>    In the meantime, I seriously suggest going to a terminal, 
>not necessarily with more capacitance, but a much bigger radius of 
>curvature. I suspect the limit has to do with the degree of output 
>ionization reaching that which imposes more-or less continuous
>loading on the secondary. I consider this to be very important, so
>much so that I am currently building a terminal for my large 
>resonator (and smaller ones) that has exactly that attribute. For 
>toroid makers out there, I am talking about the diameter of the pipe 
>from which the toroid is made, not the actual toroid diameter. The 
>limit for ROC is a sphereoid shape.
>
>Malcolm
>
>
>Malcolm,

This has always been the limiting factor!  The smallest radius of curvature
protruding from any terminal which is outside of electrostatic shielding is
the total determinant factor of break out voltage.  No other charcteristic
affects this. 

 The capacity of the terminal is simply related to the amount of energy
which may be IDEALLY stored atop the operating system.  As the energy is
1/2c(v^2), the breakout voltage, thus the smallest unshieldied radius of
curvature, can limit the energy stored regardless of capacity of the terminal.

Richard Hull, TCBOR