Re: Toroid Design .

to: Terry

The upward and downward run of large VDGRF's usually have opposite charge
on them.  They tend to attract with substantial force if wider than 12
inches.  These effects will also work to slow the drive motor somewhat.  


> From: Tesla List <tesla-at-pupman-dot-com>
> To: tesla-at-pupman-dot-com
> Subject: Re: Toroid Design .
> Date: Saturday, January 02, 1999 4:40 PM
> 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 uC
> 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 real
> 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 lectures)
> 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 
> At 06:58 AM 1/2/99 +0000, you wrote:
> >
> >  Ed -
> >
> >  The "pelletron" website was interesting but did not contain any
> >information on work regarding the VDG. I also was not able to find any
> >for VDG work on any other site or in physics and engineering texts. I
> >if anyone has ever considered this problem. Have you found info on VDGs
> >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
> >would help in undestanding the work involved with charging the Tesla
> >terminal.
> >
> >  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
> >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
> >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
> >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
> >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
> >on the belt motor. The work to charge the terminal would come from the
> >supply.
> >
> >  Determining the electrical input for the motor would be easy but
> >determining the part needed for the charging load may be impossible. It
> >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
> >text.
> >
> >  Sorry to hear about your eye problems. I also am limited by my eyes as
> >time at the computer screen.
> >
> >  John Couture