Re: Toroid Design .

"Ed -

  Isn't the  charge zero on the inside of the upper terminal during VDG
operation? How can there be an "up hill" effect for the charge on the
I would consider it a "down hill" effect.

  The work to put the charge (100% eff) on the outside of the terminal

     W = Q x V   joules per coulomb

  Have you found an equation to relate this work to the motor that would
indicate the motor slows down as the voltage on the outside of the

  John Couture"

	Yep, the charge on the INSIDE of the upper terminal IS zero, but the
VOLTAGE on the outside isn't!!!  Here is an analogy which at least hints
at what is happening.  As the belt moves "up" (toward the upper
terminal) you are moving LIKE charges to it; since like charges repel,
it takes work to move them "up", and that work is stored as additional
charge on the upper terminal.  As the voltage on the terminal increases,
the "hill" is getting higher and the repulsion is getting greater, hence
the greater load on the motor.  By the way, I built a small hand-cranked
VDG once, and it was possible to see the belt "back away" from a
highly-charged upper terminal when the crank was released.

	My eyes are on the blink at the moment, so don't feel like sitting down
at the desk and figuring, but shouldn't be much work to figure out the
actual work being done.  Will give it a try when I'm doing better.