Re: Toroid Design .
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") dia.
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
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 one
uamp per sec. to the terminal.
Charge rate = 5 x 100 = 500 sq in/50 = 10 uamps per second to the terminal.
Charge Time = 54.6/10 = 5.46 secs to charge terminal at 100% efficiency.
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 be
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 amps
at 115 volts AC. Only 10 uamps of this current may (no HV power supply) be
used to charge the terminal. This charge current is constant and independent
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 per
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 pail
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 found
the addition of charges on the inside of the pail could increase the voltage
on the outside of the pail beyond limit.
Dr. Resonance and others who build these devices may want to comment on
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 me
>explain the energy principles the best I, and Paul Tipler's College Physics
>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 an
>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 = r
>/ k Farads/meter radius)
>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 1MegV/sec
>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 so
>when the belt falls off the track, or any of a number of screw-ups happen,
>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 process
>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 the
>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 a
>"push" the belt could turn by itself in the same direction. 111 joules is
>a lot of energy so this is possible on very large systems and has been
>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 to
>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 that
>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 this
>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 it
>takes to charge them. Read the part about work and potential between point
>charges for the real theory involved. "Coulomb's Law" is the real heart of
>all this. John - If your "physics and engineering texts" don't mention ANY
>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 that
>point charges :-))