[Prev][Next][Index][Thread]

Re: Toroid Design


  Scott, Terry -

  I just noticed in your equation -

    Sphere capacitance = 4 PI EO R = 111 pf

  Isn't R the radius of the 1 meter sphere?
  The sphere capacitance would then be = 55.5 pf as I show in my post.

  The stored energy would be = 28 joules.
  Note that this is not the energy (or theoretical work) that is needed to
charge the sphere to 1 million volts. For the real world work the efficiency
would have to be known. Do you have any information on these efficiences?

  Tesla coil terminals would have similar conditions.

  John Couture


<<<< Of course, we were just testing you John :-)))  - Terry >>>>

-----------------------------------


At 08:17 PM 1/4/99 -0700, you wrote:
>Original Poster: Scott Stephens <Scott2-at-mediaone-dot-net> 
>
>At 10:02 PM 1/3/99 -0700, you wrote:
>
>According to my cypher'n:
>
>1 meter dia sphere capacitance= 4 PI E0 R = 111 pF
>Energy stored -at- 1 million volts= .5 C V^2 = 55.5 Joules
>To double the voltage on the sphere (quadruple the power to 222J)=
>222-55=167J; add 167 joules
>
>The next voltage doubling will require adding (888-222=) 666 Joules }:-)
>
>>> Original Poster: Terry Fritz <twf-at-verinet-dot-com>
>
>>> 	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) 
>>> 
>>> 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
>>1MegV/sec
>>> rate.
>
>Adding constant power will result in a limit being reached. That limit
>itself will be limited far more practicaly by the poor insulation systems
>available. I would just love to have a meter-diameter diamond or sapphire
>for an Inertial Electrostatic-confinement thermo-nuclear fusor.
>
>Forunately for Tesla-philes, the realm of dynamics offers far higher
>potentials and powers with the pulse-power paradigm.
>
>
>