Re: Toroid Design .
The "pelletron" website was interesting but did not contain any
information on work regarding the VDG. I also was not able to find any info
for VDG work on any other site or in physics and engineering texts. I wonder
if anyone has ever considered this problem. Have you found info on VDGs that
mentions work involved? It appears to be complex and to require a lot of
empirical data from real world VDGs.
I believe understanding about the work required to charge the VDG terminal
would help in undestanding the work involved with charging the Tesla coil
In your description "that as the belt moves up etc" cannot be possible
because there is no charge on the inside of the terminal to repel the charge
on the belt. No charge inside the terminal is the electrical magic that
surprised Faraday. If the belt was on the outside of the terminal the
charges would repel as you stated but that is not what happens.
You say that the "belt backed away". That could indicate a possible
repelling effect but not necessarily an increasing repelling effect with an
increasing terminal voltage.
It is obvious the work to get the charge on the VDG terminal must come
from somewhere. This work appears to be in the form of belt friction
creating electrical charges or from a separate power supply. The electrical
friction would be a load on the belt motor but would be a very small part of
the total motor load. This electrical friction load would be a constant so
would not slow up the motor as the voltage on the terminal increases. If
there is a separate power supply there would be no electrical friction load
on the belt motor. The work to charge the terminal would come from the power
Determining the electrical input for the motor would be easy but
determining the part needed for the charging load may be impossible. It is,
however, possible to determine the work to charge the terminal by Work = Q
x V where Q is in coulombs.
Note that I have not been able to verify the above VDG work theory in any
Sorry to hear about your eye problems. I also am limited by my eyes as to
time at the computer screen.
At 06:43 PM 12/31/98 -0700, you wrote:
>Original Poster: Ed Phillips <evp-at-pacbell-dot-net>
>Tesla List wrote:
>> Original Poster: Ed Phillips <evp-at-pacbell-dot-net>
>> "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
>> 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
>> 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.
>Note added 12/31/98:
> For a really neat animated illustration of charge transfer in a VDG