Re: Toroid Design .

to: John

Good post on the theory John.  One caution to watch is the leakage current
from the relatively large area of very large balls.  Point in case, we
built a 40 inch dia 1 MEV Van de Graaff several years ago and attempted to
charge it with a 6 inch wide belt running at 75 ft/sec.  It barely worked. 
Output spark was only 4 inches.  Terrible error.  With experimentation we
discovered it required a 9 inch wide belt running at 90 ft/sec which now
produces sparks around 32 inches (bright spark from side of terminal) in
length.  It nows works properly but, in theory, should have worked ok with
the first design.  The large surface area and per sq/inch leakage rate
reduced the output to under 100 kV on our first tests.  Also, note at
around 100 ft/sec or greater bearing noise becomes quite irritating.  We
prefer 65-70 ft/sec as a better design value.  Charge is also "blown" off
the belt as the charge collides with stationary air molecules on its upward

Other info for designers is, in theory, 50 uA of terminal to ground current
is produced with every sq. inch of belt passing over the pulley, however,
in actual practice 28 uA per sq. inch is a much more reliable design value.
 Output potential is usually around 68% of theorectical max (30 kV/cm
breakout value).


> From: Tesla List <tesla-at-pupman-dot-com>
> To: tesla-at-pupman-dot-com
> Subject: Re: Toroid Design .
> Date: Tuesday, January 05, 1999 6:47 AM
> Original Poster: "John H. Couture" <COUTUREJH-at-worldnet.att-dot-net> 
>   Terry, All,
>   It appears there is a big difference of opinion here. I will use your
> example to show you my solution. Your example for the one meter (39")
> sphere charged to one million volts as a VDG generator. 
>   Cap = Dia x 1.4 = 39 x 1.4 = 54.6 pf
>   Coulombs = C x V = (54.6 x 10^-12) x (10^6) = 54.6 ucoulombs to charge
> terminal.
>   A typical belt size for this VDG would be about 5" wide moving at about
> 100 inches per second. Fifty sq inches of belt per second produces about
> uamp per sec. to the terminal.
>   Charge rate = 5 x 100 = 500 sq in/50 = 10 uamps per second to the
>   Charge Time = 54.6/10 = 5.46 secs to charge terminal at 100%
>   My guess is that a separate power supply would be used because it would
> take too long to charge the terminal due to the very low efficiency.
>   My question is "has anyone ever determined the efficiency that should
> used to find the true charge time"?  I have not been able to find this
> information in the literature. Was this information in Tipler's text? 
>   If a 1/4 HP motor is used for the belt motor the current is about 5.8
> at 115 volts AC. Only 10 uamps of this current may (no HV power supply)
> used to charge the terminal. This charge current is constant and
> of the voltage on the terminal. If it varied it would not be noticed. 
>   The 10 uamps per second is a far different result than the 111 watts
> second you found which is incorrect. Your joules equation should use
> capacitance not coulombs.
>   Joules = Q x V    Joules = C x V^2    Q = C x V 
>   It is my understanding that it is not Coulomb's Law but Faraday's ice
> discovery in 1824 that makes the VDG possible. Electrical potential is a
> work function and can be cumulative. This was noted by Faraday when he
> the addition of charges on the inside of the pail could increase the
> on the outside of the pail beyond limit.
>   Dr. Resonance and others who build these devices may want to comment on
> the above.
>   John Couture
> ----------------------------------- 
> At 04:40 PM 1/2/99 -0700, you wrote:
> >Original Poster: Terry Fritz <twf-at-verinet-dot-com>
> >
> >Hi John and All,
> >
> >	Although Van de Graaff generators are a fringe topic for this list, let
> >explain the energy principles the best I, and Paul Tipler's College
> >text, can.  Also demonstrating that we just don't make this stuff up
> >
> >	Reading from Tipler's book, the work required to bring a charge Q1 from
> >infinite distance to within a radius r of another charge Q2 is:
> >
> >W = k x Q1 x Q2 / r    [joules]
> >
> >The voltage or potential of a point charge at a distance r is:
> >
> >v = k x Q / r  (since Q / v = capacitance, the capacitance of a sphere =
> >/ k  Farads/meter radius) 
> >
> >where 
> >
> >k = 1 / (4 x pi x e0)  ==~  9 x 10^9  [N-m^2/C^2]
> >
> >e0 = 8.85 x 10^-12    [ C^2/N-m^2 ] = [Farads/meter]
> >
> >So if we have a Van de Graaff sphere 1 meter in diameter charged to 1
> >million volts the charge is:
> >
> >1000000 = k Q / 1    ....    Q = 111.1 uCoulomb
> >
> >So if we want to double that voltage in 1 second, we need to add 111.1
> >to the sphere in 1 second.
> >How much power (in watts) does it take to support this charge rate?
> >
> >The work in joules is:
> >
> > k x (111.1 x 10^6)^2 / 1  which is 111.1 joules / sec or 111.1 watts if
> >the sytsem is running continuously.
> >
> >That works out to 111 watts to keep the generator charging at the
> >rate.
> >
> >Since the motor is supplying 111 watts (0.15 Horse Power) it can easily
> >slow down if it is not very strong.  Van de Graaff motors are often weak
> >when the belt falls off the track, or any of a number of screw-ups
> >the motor will do minimal damage.
> >
> >If the generator is turned off, there is a lot of charge on the top
> >terminal.  It is possible, if the belt moves freely, to reverse the
> >and use the stored energy to turn the belt in the opposite direction.
> >
> >One side of the belt is charged by the top terminal and is attracted to
> >ground.  On the other side, the belt is at ground potential and is
> >attracted to the top terminal.  So the belt can start turning.  In a
> >system the process is likely to start in the opposite direction but with
> >"push" the belt could turn by itself in the same direction.  111 joules
> >a lot of energy so this is possible on very large systems and has been
> >observed.
> >
> >Note that there is mechanical friction in this system but ideally it is
> >very low and has no real effect.  There are a few issues with how the
> >charges are created and transferred at the top and bottom of the machine
> >the belt but nothing to serious.  Needle points are used to create high
> >field stresses that arc and "suck" the charge from the belt.  The fact
> >the potential in the sphere is zero (relative to other objects in the
> >sphere) helps this process of charge transfer.  It works in reverse when
> >the machine is turned off.  The belts are electrostatically charged by
> >needle points driven by a high voltage power supply at the base.  It is
> >possible to use friction (triboelectric effect) to charge the belt but
> >method is not very good in a REAL machine. 
> >
> >Refer to any modern (Van de Graaff wasn't that old) physics book in the
> >electrostatic potential section for more details.  Many use the Van de
> >Graaff machine as an example and give homework problems on how much work
> >takes to charge them.  Read the part about work and potential between
> >charges for the real theory involved.  "Coulomb's Law" is the real heart
> >all this.  John - If your "physics and engineering texts" don't mention
> >of this, throw them away and get new ones.  The books (probably
> >Coulomb learned from, won't mention any of this either :-))
> >
> >It is interesting to note that the French physicist Charles Augustin de
> >Coulomb (1736-1806) figured all this out 214 years ago.
> >
> >Robert J. Van de Graaff was born in 1901 (young by great physicist
> >standards) but I doubt if he is still living.   
> >
> >I hope I did all this right.  In college I was attracted to girls more
> >point charges :-))
> >
> >	Terry
> >	terryf-at-verinet-dot-com